• 


RESISTANCE  OF  MATERIALS 


FOE  BEGINNERS  IN  ENGINEERING 


BY 


S.  E.  SLOCUM,  B.E.,  PH.D. 

PROFESSOR   OF  APPLIED   MATHEMATICS   IN   THE 
UNIVERSITY   OF   CINCINNATI 


GINN  AND  COMPANY 

BOSTON  •  NEW  YORK  •  CHICAGO  •  LONDON 


x  *r 

st-> 


COPYRIGHT,  1914,  BY 
S.  E.  SLOCUM 


ALL  BIGHTS  RESERVED 
414.10 


gftc  gtfttnaeum   jprrgg 

GINN  AND  COM^  \NY  •  PRO- 
PRIETORS •  BOSTON  •  U.S.A. 


PREFACE 

The  chief  feature  which  distinguishes  this  volume  from  other 
American  textbooks  on  the  same  subject  is  that  the  Principle  of 
Moments  is  used  consistently  throughout  in  place  of  the  usual 
calculus  processes.  By  basing  the  work  on  this  principle  it  has 
been  found  practicable  to  give  a  simple  and  obvious  treatment 
of  many  topics  for  which  the  calculus  is  usually  thought  to  be 
indispensable,  such  as  the  calculation  of  moments  of  inertia,  the 
deflection  of  beams,  the  buckling  of  columns,  and  the  strength 
of  thick  cylinders.  Experience  has  shown  conclusively  that  the 
average  engineering  graduate,  and  even  the  practicing  engineer,  is 
deficient  in  the  ability  to  apply  the  Principle  of  Moments  readily, 
but  when  thus  used  as  the  central  and  coordinating  principle,  it 
must  necessarily  make  an  indelible  impression  on  the  mind  of  the 
student  and  go  far  toward  remedying  this  deficiency. 

The  mechanics  of  materials  is  of  such  fundamental  importance  in 
all  branches  of  technology  that  it  is  important  to  begin  its  study 
as  early  in  the  course  as  possible.  Heretofore  it  has  been  necessary 
to  defer  it  -  -  awaiting  the  completion  of  the  calculus  — until  junior 
year,  when  the  curriculum  is  already  crowded  with  technical  sub- 
jects requiring  its  application.  This  text  makes  it  possible  for  the 
course  to  parallel  or  even  to  precede  the  calculus.  In  addition,  it 
makes  the  subject  available  for  trade  or  architectural  schools  where 
no  calculus  is  taught. 

Although  simple  and  obvious,  the  treatment  is  adequate,  and 
its  simplicity  in  no  way  limits  its  range  or  generality.  The  text 
is  supplemented  by  a  variety  of  engineering  applications,  giving 
practical  information  as  well  as  a  mastery  of  the  principles  involved. 

S.  E.  SLOCUM 


CONTENTS 

SECTION  I 
STRESS  AND  DEFORMATION 

PAGES 

Elastic  resistance,  or  stress.  —  Varieties  of  strain.  —  Strain  diagram. 
—  Hooke's  law.  —  Elastic  limit.  —  Working  stress.  —  Resilience.  — 
Poisson's  ratio.  —  Temperature  stress.  —  Applications 1-14 

SECTION  II 
FIRST  AND  SECOND  MOMENTS 

Static  moment.  —  Fundamental  theorem  of  moments.  —  Center  of 
gravity,  -r-  Centroid.  —  Centroid  of  triangular  area.  —  Centroid  of 
circular  arc.  —  Centroid  of  circular  sector  and  segment.  —  Centroid 
of  parabolic  segment.  —  Axis  of  symmetry.  —  Centroid  of  composite 
figures.  —  Moment  of  inertia.  —  I  for  rectangle.  —  I  for  triangle.  — 
I  for  circle.  —  I  for  composite  figures.  —  Applications 15-34 

SECTION  III 
BENDING-MOMENT  AND  SHEAR  DIAGRAMS 

Conditions  of  equilibrium. — Vertical  shear.  —  Bending  moment. 

—  Bending-moment  and  shear  diagrams.  —  Relation  between  shear 
and  moment  diagrams.  —  Properties  of  shear  and  moment  diagrams. 

—  General  directions  for  sketching  diagrams.  —  Applications       .     .       35~48 

SECTION  IV 
STRENGTH  OF  BEAMS 

Nature  of  bending  stress.  —  Distribution  of  stress.  —  Fundamental 
formula  for  beams.  —  Calculation  and  design  of  beams. — Applications  49-59 


vi  CONTENTS 

SECTION  V 
DEFLECTION  OF  CANTILEVER  AND  SIMPLE  BEAMS 

PAGES 

General  deflection  formula.  —  Cantilever  bearing  concentrated 
load.  —  Cantilever  bearing  uniform  load.  —  Cantilever  under  constant 
moment.  —  Simple  beam  bearing  concentrated  load.  —  Simple  beam 
bearing  uniform  load.  —  Applications 60-69 

SECTION  VI 
CONTINUOUS  BEAMS 

Theorem  of  three  moments  for  uniform  loads.  —  Theorem  of  three 
moments  for  concentrated  loads.  —  Effect  of  unequal  settlement  of 
supports.  —  Applications 70-79 

SECTION  VII 
RESTRAINED,  OR  BUILT-IN,  BEAMS 

Uniformly  loaded  beam  fixed  at  both  ends.  —  Beam  fixed  at  both 
ends  and  bearing  concentrated  load  at  center.  —  Single  eccentric  load. 
—  Uniformly  loaded  beam  fixed  at  one  end.  —  Beam  fixed  at  one  end 
and  bearing  concentrated  load  at  center.  —  Beam  fixed  at  one  end  and 
bearing  a  concentrated  eccentric  load.  —  Applications 80-90 

SECTION  VIII 

COLUMNS  AND  STRUTS 

Nature  of  compressive  stress.  —  Euler's  theory  of  long  columns.  — 
Effect  of  end  support.  —  Modification  of  Euler's  formula.  —  Ran- 
kine's  formula.  —  Values  of  the  empirical  constants  in  Rankine's 
formula.  —  Johnson's  parabolic  formula.  —  Johnson's  straight-line 
formula.  —  Cooper's  modification  of  Johnson's  straight-line  formula. 
—  Eccentrically  loaded  columns.  —  Applications 91-105 

SECTION  IX 
TORSION 

Maximum  stress  in  circular  shafts.  —  Angle  of  twist  in  circular 
shafts.  —  Power  transmitted  by  circular  shafts.  —  Combined  bending 
and  torsion.  —  Resilience  of  circular  shafts.  —  Non-circular  shafts.  — 
Elliptical  shaft.  —  Rectangular  and  square  shafts.  —  Triangular 
shafts.  Angle  of  twist  for  shafts  in  general.  —  Applications  .  .  106-117 


CONTENTS  vii 

SECTION  X 
SPHERES  AND  CYLINDERS  UNDER  UNIFORM  PRESSURE 

PAGES 

Hoop  stress.  —  Hoop  tension  in  hollow  sphere.  —  Hoop  tension  in 
hollow  circular  cylinder.  —  Longitudinal  stress  in  hollow  circular 
cylinder.  — Thick  cylinders.  —  Lamp's  formulas.  —  Maximum  stress 
in  thick  cylinder  under  uniform  internal  pressure.  —  Bursting  pres- 
sure for  thick  cylinder.  —  Maximum  stress  in  thick  cylinder  under 
uniform  external  pressure.  — Comparison  of  formulas  for  the  strength 
of  tubes  under  uniform  internal  pressure.  —  Thick  cylinders  built  up 
of  concentric  tubes.  —  Practical  formulas  for  the  collapse  of  tubes 
under  external  pressure.  —  Shrinkage  and  forced  fits.  —  Applications  118-135 

SECTION  XI 

FLAT  PLATES 

Theory  of  flat  plates.  —  Maximum  stress  in  homogeneous  circu- 
lar plate  under  uniform  load.  —  Maximum  stress  in  homogeneous 
circular  plate  under  concentrated  load.  —  Dangerous  section  of  ellip- 
tical plate.  —  Maximum  stress  in  homogeneous  elliptical  plate  under 
uniform  load.  —  Maximum  stress  in  homogeneous  square  plate  under 
uniform  load.  —  Maximum  stress  in  homogeneous  rectangular  plate 
under  uniform  load.  —  Applications 136-145 

SECTION  XII 

RIVETED  JOINTS  AND  CONNECTIONS 

Efficiency  of  riveted  joint.  —  Boiler  shells.  —  Structural  steel.  — 
Unit  stresses.  —  Applications 146-155 

SECTION  XIII 

REENFORCED  CONCRETE 

Physical  properties.  —  Design  of  reenforced-concrete  beams.— 
Calculation  of  stirrups,  or  web  reinforcement.  —  Reenforced- 
concrete  columns.  —  Radially  reenf orced  flat  slabs.  —  Diameter  of 
top.  —  Efficiency  of  the  spider  hoops.  —  Maximum  moment. —  Thick- 
ness of  slab.  —  Area  of  slab  rods.  —  Application  of  formulas.  — 
Dimension  table.  —  Dimensions  of  spider.  —  Applications.  .  .  .  156-175 


viii  CONTENTS 

SECTION  XIV 
SIMPLE   STRUCTURES 

PAGES 

Composition  and  resolution  of  forces Conditions  of  equilibrium 

of  a  system  of  coplanar  forces.  —  Equilibrium  polygon.  —  Applica- 
tion of  equilibrium  polygon  to  determining  reactions.  —  Equilibrium 
polygon  through  two  given  points.  —  Equilibrium  polygon  through 
three  given  points.  —  Application  of  equilibrium  polygon  to  calcula- 
tion of  stresses.  —  Relation  of  equilibrium  polygon  to  bending- 
moment  diagram.  —  Structures  :  external  forces.  —  Structures  :  joint 
reactions.  —  Structures  :  method  of  sections.  —  Applications  .  .  .  176-204 

ANSWERS  TO  PROBLEMS 205-207 

INDEX .  209-210 


TABLES  OF  PHYSICAL  AND  MATHEMATICAL 

CONSTANTS 

I.  AVERAGE  VALUES  OF  PHYSICAL  CONSTANTS 

II.  PROPERTIES  OF  VARIOUS  SECTIONS 

III.  PROPERTIES  OF  STANDARD  I-BEAMS 

IV.  PROPERTIES  OF  STANDARD  CHANNELS 
V.  PROPERTIES  OF  STANDARD  ANGLES 

VI.    PROPERTIES  OF  BETHLEHEM  GIRDER  BEAMS 
VII.    PROPERTIES  OF  BETHLEHEM  I-BEAMS 
VIII.   MOMENTS  OF  INERTIA  AND  SECTION  MODULI  :  RECTANGULAR  CROSS 

SECTION 
IX.    MOMENTS    OF    INERTIA  AND    SECTION   MODULI  :    CIRCULAR   CROSS 

SECTION 

X.    FOUR-PLACE  LOGARITHMS  OF  NUMBERS 
XI.    CONVERSION  OF  LOGARITHMS 
XII.    FUNCTIONS  OF  ANGLES 

XIII.  BENDING-MOMENT  AND  SHEAR  DIAGRAMS 

XIV.  MENSURATION 

XV.    FRACTIONAL  AND  DECIMAL  EQUIVALENTS 
XVI.    WEIGHTS  OF  VARIOUS  SUBSTANCES 
XVII.    STRENGTH  OF  ROPES  AND  BELTS 


ix 


TABLES 


XI 


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RESISTANCE  OF  MATERIALS 


TABLE  I  (Continued) 
2.   POISSON'S  RATIO 


MATERIAL 

AVERAGE 
VALUES  OF 
1 

m 

Steel   hard 

295 

"      structural 

299 

.277 

Brass    .          .... 

357 

Copper 

340 

Lead     

.375 

Zinc      .                

.205 

3.    FACTORS  OF  SAFETY 


STEADY 

VARYING 

REPEATED  OR 

MATERIAL 

STRESS  : 
BUILDINGS, 

STRESS  : 
BRIDGES, 

REVERSED 
STRESS  : 

ETC. 

ETC. 

MACHINES 

Steel,  hard     

5 

8 

15 

"      structural      

4 

6 

10 

Iron   wrought     ... 

4 

6 

10 

"     cast 

6 

10 

20 

Timber      

8 

10 

15 

Brick  and  stone                 .... 

15 

25 

30 

The  only  rational  method  of  determining  the  factor  of  safety  is  to  choose  it 
sufficiently  large  to  bring  the  working  stress  well  within  the  elastic  limit  (see 
Article  6). 


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XV1U 


RESISTANCE  OF  MATERIALS 


TABLE  III 
PROPERTIES  OF  STANDARD  I-BEAMS 


DEPTH 

OF 

BEAM 

WEIGHT 

PER 

FOOT 

AREA 

OF 

SECTION 

THICK- 
NESS OF 
WEB 

WIDTH 

OF 

FLANGE 

MOMENT 

OF 

INERTIA 
Axis  1-1 

SECTION 
MODU- 
LUS 
Axis  1-1 

RADIUS 

OF 

GYRA- 
TION 
Axis  1-1 

MOMENT 

OF 

INERTIA 
Axis  2-2 

RADIUS 

OF 

GYRA- 
TION 
Axis  2-2 

d 

A 

t 

b 

I 

S 

r 

I' 

r' 

Inches 

Pounds 

Sq.  Inches 

Inches 

Inches 

Inches* 

Inches  & 

Inches 

Inches  * 

Inches 

3 

55 

1.63 

.17 

2.33 

2.5 

1.7 

1.23 

.46 

53 

ii 

6.5 

1.91 

.26 

2.42 

2.7 

1.8 

1.19 

53 

52 

" 

7.5 

2.21 

.36 

252 

2.9 

1.9 

1.15 

.60 

52 

4 

75 

2.21 

.19 

2.66 

6.0 

3.0 

1.64 

.77 

59 

85 

2.50 

.26 

2.73 

6.4 

3.2 

1.59 

.85 

58 

II 

9.5 

2.79 

.34 

2.81 

6.7 

3.4 

154 

.93 

58 

II 

105 

3.09 

.41 

2.88 

7.1 

3.6 

152 

1.01 

57 

5 

9.75 

2.87 

21 

3.00 

12.1 

4.8 

2.05 

1.23 

.65 

ii 

12.25 

3.60 

!36 

3.15 

13.6 

5.4 

1.94 

1.45 

.63 

" 

14.75 

4.34 

50 

3.29 

15.1 

6.1 

1.87 

1.70 

.63 

6 

12.25 

3.61 

.23 

3.33 

21.8 

7.3 

2.46 

1.85 

.72 

14.75 

4.34 

.35 

3.45 

24.0 

8.0 

2.35 

2.09 

.69 

"    . 

17.25 

5.07 

.47 

357 

26.2 

8.7 

2.27 

2.36 

.68 

7 

15.0 

4.42 

.25 

3.66 

36.2 

10.4 

2.86 

2.67 

.78 

175 

5.15 

.35 

3.76 

39.2 

11.2 

2.76 

2.94 

.76 

ii 

20.0 

5.88 

.46 

3.87 

42.2 

12.1 

2.68 

3.24 

.74 

8 

17.75 

5.33 

.27 

4.00 

56.9 

14.2 

3.27 

3.78 

.84 

•i 

20.25 

5.96 

.35 

4.08 

60.2 

15.0 

3.18 

4.04 

.82 

it 

22.75 

6.69 

.44 

4.17 

64.1 

16.0 

3.10 

4.36 

.81 

" 

25.25 

7.43 

.53 

4.26 

68.0 

17.0 

3.03 

4.71 

.80 

0 

21.0 

6.31 

.29 

4.33 

84.9 

18.9 

3.67 

5.16 

.90 

ii 

25.0 

7.35 

.41 

4.45 

91.9 

20.4 

3.54 

5.65 

.88 

ii 

30.0 

8.82 

57 

4.61 

101.9 

22.6 

3.40 

6.42 

.85 

" 

35.0 

10.29 

.73 

4.77 

111.8 

24.8 

3.30 

7.31 

.84 

10 

25.0 

7.37 

.31 

4.66 

122.1 

24.4 

4.07 

6.89 

57 

II 

30.0 

8.82 

.45 

4.80 

134.2 

26.8 

3.90 

7.65 

.93 

" 

35.0 

10.29 

.60 

4.95 

146.4 

29.3 

3.77 

852 

51 

" 

40.0 

11.76 

.75 

5.10 

158.7 

31.7 

3.67 

950 

50 

12 

325 

9.26 

.35 

5.00 

215.8 

36.0 

4.83 

950 

1.01 

" 

35.0 

10.29 

.44 

5.09 

228.3 

38.0 

4.71 

10.07 

.99 

" 

40.0 

11.76 

56 

5.21 

245.9 

41.0 

457 

10.95 

56 

15 

42.0 

12.48 

.41 

550 

441.8 

58.9 

5.95 

14.62 

1.08 

" 

45.0 

13.24 

.46 

555 

455.8 

60.8 

5.87 

15.09 

1.07 

«< 

50.0 

14.71 

56 

5.65 

483.4 

645 

5.73 

16.04 

1.04 

ii 

55.0 

16.18 

.66 

5.75 

511.0 

68.1 

5.62 

17.06 

1.03 

" 

60.0 

17.65 

.75 

5.84 

538.6 

71.8 

552 

18.17 

1.01 

18 

55.0 

15.93 

.46 

6.00 

795.6 

88.4 

7.07 

21.19 

1.15 

«« 

60.0 

17.65 

56 

6.10 

841.8 

935 

6.91 

22.38 

1.13 

ii 

65.0 

19.12 

.64 

6.18 

8815 

97.9 

6.79 

23.47 

1.11 

" 

70.0 

20.59 

.72 

6.26 

921.2 

102.4 

6.69 

24.62 

1.09 

20 

65.0 

19.08 

50 

6.25 

11695 

117.0 

7.83 

27.86 

1.21 

" 

70.0 

20.59 

58 

6.33 

1219.8 

122.0 

7.70 

29.04 

1.19 

" 

75.0 

22.06 

.65 

6.40 

1268.8 

126.9 

758 

30.25 

1.17 

24 

80.0 

23.32 

.50 

7.00 

2087.2 

173.9 

9.46 

42.86 

1.36 

i* 

85.0 

25.00 

57 

7.07 

2167.8 

180.7 

9.31 

44.35 

1.33 

" 

90.0 

26.47 

.63 

7.13 

2238.4 

1865 

9.20 

45.70 

1.31 

" 

95.0 

27.94 

.69 

7.19 

2309.0 

192.4 

9.09 

77.10 

1.30 

100.0 

29.41 

.75 

7.25 

2379.6 

198.3 

859 

4855 

1.28 

TABLES 

TABLE  IV 
PROPERTIES  OF  STANDARD  CHANNELS 


XIX 


.(• a-      ff  ^i  ^ 


s 

i 

03 

,   ^ 

<N 

09 

. 

JH 

DEPTH  OF 
CHANNEL 

WEIGHT  PER 
FOOT 

AREA  OF  SECTK 

THICKNESS  OF 
WEB 

WIDTH  OF 
FLANGE 

MOMENT  OF 
INERTIA  Axis  1 

I 

SECTION  MODUL 
Axis  1-1 

RADIUS  OF  GYR 
TION  AXIS  1-1 

MOMENT  OF 
INERTIA  Axis  2 

SECTION  MODUL 
Axis  2-2 

RADIUS  OF  GYR 
TION  Axis  2-2 

DISTANCE  OF 
CENTER  OF  GRA 
ITY  FROM  OUTSIJ 
OF  WEB 

d 

A 

t 

b 

I 

S 

r 

r 

S' 

r' 

X 

Inches 

Pounds 

Sq.In. 

Inches 

Inches 

Inches* 

Inches3 

Inches 

Inches* 

Inches3 

Inches 

Inches 

3 

4.00 

1.19 

.17 

1.41 

1.6 

1.1 

1.17 

.20 

.21 

.41 

-.44 

«« 

5.00 

1.47 

.26 

1.50 

1.8 

1.2 

1.12 

.25 

.24 

.41 

.44 

" 

6.00 

1.76 

.36 

1.60 

2.1 

1.4 

1.08 

.31 

.27 

.42 

.46 

4 

5.25 

1.55 

.18 

158 

3.8 

1.9 

156 

.32 

.29 

.45 

.46 

«« 

6.25 

1.84 

.25 

1.65 

4.2 

2.1 

1.51 

.38 

.32 

.45 

.46 

" 

7.25 

2.13 

.33 

1.73 

4.6 

2.3 

1.46 

.44 

.35 

.46 

.46 

5 

6.50 

1.95 

.19 

1.75 

7.4 

3.0 

1.95 

.48 

.38 

50 

.49 

" 

9.00 

2.65 

.33 

1.89 

8.9 

3.5 

1.83 

.64 

.45 

.49 

.48 

it 

11.50 

3.38 

.48 

2.04 

10.4 

4.2 

1.75 

.82 

.54 

.49 

51 

6 

8.00 

2.38 

.20 

1.92 

13.0 

4.3 

2.34 

.70 

50 

.54 

52 

" 

1050 

3.09 

.32 

2.04 

15.1 

5.0 

2.21 

.88 

57 

.53 

50 

" 

13.00 

3.82 

.44 

2.16 

17.3 

5.8 

2.13 

1.07 

.65 

53 

52 

" 

15.50 

4.56 

56 

2.28 

19.5 

6.5 

2.07 

1.28 

.74 

.53 

55 

7 

9.75 

2.85 

.21 

2.09 

21.1 

6.0 

2  72 

.98 

.63 

59 

.55 

«< 

12.25 

3.60 

.32 

2.20 

24.2 

6.9 

259 

1.19 

.71 

57 

53 

«« 

14.75 

4.34 

.42 

2.30 

27.2 

7.8 

2.50 

1.40 

.79 

.57 

53 

" 

17.25 

5.07 

.53 

2.41 

30.2 

8.6 

2.44 

1.62 

.87 

56 

55 

" 

19.75 

5.81 

.63 

2.51 

33.2 

9.5 

2.39 

1.85 

.96 

56 

58 

8 

11.25 

3.35 

.22 

2.26 

32.3 

8.1 

3.10 

1.33 

.79 

.63 

58 

" 

13.75 

4.04 

.31 

2.35 

30.0 

9.0 

2.98 

1.55 

.87 

.62 

56 

«< 

16.25 

4.78 

.40 

2.44 

39.9 

10.0 

2.89 

1.78 

.95 

.61 

.56 

«« 

18.75 

551 

.49 

2.53 

43.8 

11.0 

2.82 

2.01 

1.02 

.60 

.57 

" 

21.25 

6.25 

2.62 

47.8 

11.9 

2.76 

2.25 

1.11 

.60 

59 

9 

13.25 

3.89 

.23 

2.43 

47.3 

10.5 

3.49 

1.77 

.97 

.67 

.61 

" 

15.00 

4.41 

.29 

2.49 

50.9 

11.3 

3.40 

1.95 

1.03 

.66 

59 

" 

20.00 

5.88 

.45 

2.65 

60.8 

13.5 

3.21 

2.45 

1.19 

.65 

58 

" 

25.00 

7.35 

.61 

2.81 

70.7 

15.7 

3.10 

2.98 

1.36 

.64 

.62 

10 

15.00 

4.46 

.24 

2.60 

66.9 

13.4 

3.87 

2.30 

1.17 

-.72 

.64 

" 

20.00 

5.88 

.38 

2.74 

78.7 

15.7 

3.66 

2.85 

1.34 

.70 

.61 

» 

25.00 

7.35 

53 

2.89 

91.0 

18.2 

3.52 

3.40 

1.50 

.68 

.62 

" 

30.00 

8.82 

.68 

3.04 

103.2 

20.6 

3.42 

3.99 

1.67 

.67 

.65 

" 

35.00 

10.29 

.82 

3.18 

115.5 

23.1 

3.35 

4.66 

1.87 

.67 

.69 

12 

20.50 

6.03 

.28 

2.94 

128.1 

21.4 

4.61 

3.91 

1.75 

.81 

.70 

«« 

25.00 

7.35 

.39 

3.05 

144.0 

24.0 

4.43 

4.53 

1.91 

.78 

.68 

«« 

30.00 

8.82 

51 

3.17 

161.6 

26.9 

4.28 

5.21 

2.09 

.77 

.68 

" 

35.00 

10.29 

.64 

3.30 

179.3 

29.9 

4.17 

5.90 

2.27 

.76 

.69 

" 

40.00 

11.76 

.76 

3.42 

196.9 

32.8 

4.09 

6.63 

2.46 

.75 

.72 

15 

33.00 

9.90 

.40 

3.40 

312.6 

41.7 

5.62 

8.23 

3.16 

.91 

.79 

«< 

35.00 

10.29 

.43 

3.43 

319.9 

42.7 

5.57 

8.48 

3.22 

.91 

.79 

« 

40.00 

11.76 

3.52 

347.5 

46.3 

5.44 

9.39 

3.43 

.89 

.78 

" 

45.00 

13.24 

.62 

3.62 

375.1 

50.0 

5.32 

10.29 

3.63 

.88 

.79 

» 

50.00 

14.71 

.72 

3.72 

402.7 

53.7 

5.23 

11.22 

3.85 

.87 

.80 

" 

55.00 

16.18 

.82 

3.82 

430.2 

57.4 

5.16 

12.19 

4.07 

.87 

.82 

XX 


RESISTANCE  OF  MATERIALS 


TABLE  V 
PROPERTIES  OF  STANDARD  ANGLES,  EQUAL  LEGS 


< 

\ 

1 

•1 

, 

\ 

\                 j 

1 

£ 

<  \      1 

\  t      1 

"V-^^v*! 
\/^ 

DIMENSIONS 

THICKNESS 

WEIGHT  PER  FOOT 

AREA  OF  SECTION 

DISTANCE  OF  CENTER 
OF  GRAVITY  FROM 
BACK  OF  FLANGE 

MOMENT  OF  INERTIA 
Axis  1-1 

SECTION  MODULUS 
Axis  1-1 

: 

RADIUS  OF  GYRA- 
TION Axis  1-1 

DISTANCE  OF  CENTER 
OF  GRAVITY  FROM 
EXTERNAL  APEX  ON 
LINE  INCLINED  AT 
45°  TO  FLANGE 

LEAST  MOMENT  OF 
INERTIA  Axis  2-2 

SECTION  MODULUS 
Axis  2-2 

LEAST  RADIUS  OF 
GYRATION  Axis  2-2 

Inches 

Inches 

Pounds 

Sg.In. 

Inches 

Inches* 

Inches3 

Inches 

Inches 

Inches* 

Inches5 

Inches 

fx    2 

k 

.58 

.17 

.23 

.009 

.017 

.22 

.33 

.004 

.011 

.14 

1    x  1 

i 

.80 

.23 

.30 

.022 

.031 

.30 

.42 

.009 

.021 

.19 

" 

I 

1.49 

.44 

.34 

.037 

.056 

.29 

.48 

.016 

.034 

.19 

li  x  1J 

i 

1.02 

.30 

.36 

.044 

.049 

.38 

.51 

.018 

.035 

.24 

i 

1.91 

.56 

.40 

.077 

.091 

.37 

.57 

.033 

.057 

.24 

**  «  u 

1 

2.34 

.69 

.47 

.14 

.134 

.45 

.66 

.058 

.088 

.29 

i 

3.35 

.98 

.51 

.19 

.188 

.44 

•72 

.082 

.114 

.29 

1|  X  1| 

j 

2.77 

.81 

.53 

.23 

.19 

.53 

.75 

.094 

.13 

.34 

3.98 

1.17 

57 

.31 

.26 

.51 

.81 

.133 

.16 

.34 

2x2 

i 

3.19 

.94 

.59 

.35 

.25 

.61 

.84 

.14 

.17 

.39 

" 

i 

4.62 

1.36 

.64 

.48 

.35 

.59 

.90 

.20 

.22 

.39 

2*  x  2£ 

4.0 

1.19 

.72 

.70 

.39 

.77 

1.01 

.29 

.28 

.49 

5.9 

1.73 

.76 

.98 

.57 

.75 

1.08 

.41 

.38 

.48 

" 

7.7 

2.25 

.81 

1.23 

.72 

.74 

1.14 

52 

.46 

.48 

3    X3 

4.9 

1.44 

.84 

1.24 

.58 

.93 

1.19 

.50 

.42 

.59 

" 

7.2 

2.11 

.89 

1.76 

.83 

.91 

1.26 

.72 

57 

.58 

" 

^ 

9.4 

2.75 

.93 

2.22 

1.07 

.90 

1.32 

.92 

.70 

58 

" 

i 

11.4 

3.36 

.98 

2^62 

1.30 

.88 

1.38 

1.12 

.81 

58 

3*  x  3£ 

i 

8.4 

2.48 

1.01 

2.87 

1.15 

1.07 

1.43 

1.16 

.81 

.68 

^ 

11.1 

3.25 

1.06 

3.64 

1.49 

1.06 

1.50 

1.50 

1.00 

.68 

" 

§ 

13.5 

3.98 

1.10 

4.33 

1.81 

1.04 

1.56 

1.82 

1.17 

.68 

" 

2 

15.9 

4.69 

1.15 

4.96 

2.11 

1.03 

1.62 

2.13 

1.31 

.67 

i    x4 

3 

9.7 

2.86 

1.14 

4.36 

1.52 

1.23 

1.61 

1.77 

1.10 

.79 

«< 

1 

12.8 

3.75 

1.18 

5.56 

1.97 

1.22 

1.67 

2.28 

1.36 

.78 

«< 

| 

15.7 

4.61 

1.23 

6.66 

2.40 

1.20 

1.74 

2.76 

1.59 

.77 

" 

2 

18.5 

5.44 

1.27 

7.66 

2.81 

1.19 

1.80 

3.23 

1.80 

.77 

6x6 

1 

19.6 

5.75 

1.68 

19.91 

4.61 

1.86 

2.38 

8.04 

3.37 

1.18 

" 

§ 

24.2 

7.11 

1.73 

24.16 

5.66 

1.84 

2.45 

9.81 

4.01 

1.17 

«« 

a 

28.7 

8.44 

1.78 

28.15 

6.66 

1.83 

2.51 

11.52 

4.59 

1.17 

" 

j 

33.1 

9.73 

1.82 

31.92 

7.63 

1.81 

2.57 

13.17 

5.12 

1.16 

TABLES 


XXI 


TABLE  V 
PROPERTIES  OF  STANDARD  ANGLES,  UNEQUAL  LEGS 


W 

•«* 

65 

DIMENSIONS 

THICKNESS 

WEIGHT  PER  FOOT 

AREA  OF  SECTION 

DISTANCE  OF  CENTE 
OF  GRAVITY  FROM 
BACK  OF  LONGER 
FLANGE 

MOMENT  OF  INERTI. 
Axis  1-1 

SECTION  MODULUS 
Axis  1-1 

RADIUS  OF  GYRA- 
TION Axis  1-1 

DISTANCE  OF  CENTE 
OF  GRAVITY  FROM 
BACK  OF  SHORTER 
FLANGE 

MOMENT  OF  INERTI. 
Axis  2-2 

SECTION  MODULUS 
Axis  2-2 

RADIUS  OF  GYRA- 
TION Axis  2-2 

Inches 

Inches 

Pounds 

Sq.In. 

Inches 

Inches* 

Inches^ 

Inches 

Inches 

Inches* 

Inches3 

Inches 

2$  x  2 

3.6 

1.06 

.54 

.37 

.25 

.59 

.79 

.65 

.38 

.78 

" 

5.3 

1.55 

.58 

.51 

.36 

.58 

.83 

.91 

.55 

.77 

" 

6.8 

2.00 

.63 

.64 

.46 

£6 

.88 

1.14 

.70 

.75 

3    x  2J 

4.5 

1.31 

.66 

.74 

.40 

75 

.91 

1.17 

.56 

.95 

" 

6.5 

1.92 

.71 

1.04 

.58 

.74 

.96 

1.66 

.81 

.93 

" 

8.5 

2.50 

•75 

1.30 

.74 

.72 

1.00 

2.08 

1.04 

.91 

3$  x  2J 

4.9 

1.44 

.61 

.78 

.41 

.74 

1.11 

1.80 

.75 

1.12 

" 

7.2 

2.11 

.66 

1.09 

.59 

.72 

1.16 

2.56 

1.09 

1.10 

«« 

9.4 

2.75 

.70 

1.36 

.76 

.70 

1.20 

3.24 

1.41 

1.09 

" 

11.4 

3.36 

.75 

1.61 

.92 

.69 

1.25 

3.85 

1.71 

1.07 

3$  x  3 

| 

7.8 

2.30 

.83 

1.85 

.85 

.90 

1.08 

2.72 

1.13 

1.09 

" 

i 

10.2 

3.00 

.88 

2.33 

1.10 

.88 

1.13 

3.45 

1.45 

1.07 

" 

1 

12.5 

3.67 

.92 

2.76 

1.33 

.87 

1.17 

4.11 

1.76 

1.06 

" 

1 

14.7 

4.31 

.96 

3.15 

1.54 

.85 

1.21 

4.70 

2.05 

1.04 

4x3 

1 

8.5 

2.48 

.78 

1.92 

.87 

.88 

1.28 

3.96 

1.46 

1.26 

«« 

11.1 

3.25 

.83 

2.42 

1.12 

.86 

1.33 

5.05 

1.89 

1.25 

" 

13.6 

3.98 

.87 

2.87 

1.35 

.85 

1.37 

6.03 

2.30 

1.23 

" 

15.9 

4.69 

.92 

3.28 

1.57 

.84 

1.42 

6.93 

2.68 

1.22 

5x3 

9.7 

2.86 

.70 

2.04 

.89 

.84 

1.70 

7.37 

2.24 

1.61 

" 

12.8 

3.75 

.75 

2.58 

1.15 

.83 

1.75 

9.45 

2.91 

1.59 

" 

15.7 

4.61 

.80 

3.06 

1.39 

.82 

1.80 

11.37 

3.55 

1.57 

" 

2 

18.5 

5.44 

.84 

3.51 

1.62 

.80 

1.84 

13.15 

4.16 

1.55 

5    x  3£ 

10.4 

3.05 

.86 

3.18 

1.21 

1.02 

1.61 

7.78 

2.29 

1.60 

" 

13.6 

4.00 

.91 

4.05 

1.56 

1.01 

1.66 

9.99 

2  99 

1.58 

" 

t 

16.7 

4.92 

.95 

4.83 

1.90 

.99 

1.70 

12.03 

3.65 

1.56 

» 

a 

19.8 

5.81 

1.00 

5.55 

2.22 

.98 

1.75 

13.92 

4.28 

1.55 

" 

i 

22.7 

6.67 

1.04 

6.21 

2.52 

.96 

1.79 

15.67 

4.88 

1.53 

6x3$ 

11.6 

3.42 

.79 

3.34 

1.23 

.99 

2.04 

12.86 

3.24 

1.94 

«« 

15.3 

4.50 

.83 

4.25 

1.59 

.97 

2.08 

16.59 

4.24 

1.92 

» 

18.9 

5.55 

.88 

5.08 

1.94 

.96 

2.13 

20.08 

5.19 

1.90 

«' 

22.3 

6.56 

.93 

5.84 

2.27 

.94 

2.18 

23.34 

6.10 

1.89 

" 

25.7 

7.55 

.97 

6.55 

2.59 

.93 

2.22 

26.39 

6.98 

1.87 

6x4 

12.3 

3.61 

.94 

4.90 

1.60 

1.17 

1.94 

13.47 

3.32 

1.93 

» 

16.2 

4.75 

.99 

6.27 

2.08 

1.15 

1.99 

17.40 

4.33 

1.91 

«< 

19.9 

5.86 

1.03 

7.52 

2.54 

1.13 

2.03 

21.07 

5.31 

1.90 

«« 

23.6 

6.94 

1.08 

8.68 

2.97 

1.12 

2.08 

24.51 

6.25 

1.88 

" 

27.2 

7.98 

1.12 

9.75 

3.39 

1.11 

2.12 

27.73 

7.15 

1.86 

XX11 


RESISTANCE  OF  MATERIALS 


TABLE  VI 
PROPERTIES  OF  BETHLEHEM  GIRDER  BEAMS 


DEPTH 

WEIGHT 

AREA 

THICK- 

WIDTH 

NEUTRAL  Axis  PERPEN- 

NEUTRAL Axis 
COINCIDENT  WITH 

OF 

BEAM 

PER 

FOOT 

OF 

SECTION 

NESS  OF 
WEB 

OF 

FLANGE 

DICULAR  TO  WEB  AT 
CENTER 

CENTER  LINE 
OF  WEB 

Inches 

Pounds 

Square 
Inches 

Inches 

Inches 

Moment 
of 
Inertia 

Radius 
of 
Gyration 

Section 
Modulus 

Moment 
of 
Inertia 

Radius 
of 
Gyration 

30 

200.0 

58.71 

.750 

15.00 

9150.6 

12.48 

610.0 

630.2 

3.28 

u 

180.0 

53.00 

.690 

13.00 

8194.5 

12.43 

546.3 

433.3 

2.86 

28 

180.0 

52.86 

.690 

14.35 

7264.7 

11.72 

518.9 

533.3 

3.18 

K 

165.0 

48.47 

.660 

12.50 

6562.7 

11.64 

468.8 

371.9 

2.77 

26 

160.0 

46.91 

.630 

13.60 

5620.8 

10.95 

432.4 

435.7 

3.05 

u 

150.0 

43.94 

.630 

12.00 

5153.9 

10.83 

396.5 

314.6 

2.68 

24 

140.0 

41.16 

.600 

13.00 

4201.4 

10.10 

350.1 

346.9 

2.90 

« 

120.0 

35.38 

.530 

12.00 

3607.3 

10.10 

300.6 

249.4 

2.66 

20 

140.0 

41.19 

.640 

12.50 

2934.7 

8.44 

293.5 

348.9 

2.91 

« 

112.0 

32.81 

.550 

12.00 

2342.1 

8.45 

234.2 

239.3 

2.70 

18 

92.0 

27.12 

.480 

11.50 

1591.4 

7.66 

176.8 

182.6 

2.59 

15 

140.0 

41.27 

.800 

11.75 

1592.7 

6.21 

212.4 

331.0 

2.83 

« 

104.0 

30.50 

.600 

11.25 

1220.1 

6.32 

162.7 

213.0 

2.64 

u 

73.0 

21.49 

.430 

10.50 

883.4 

6.41 

117.8 

123.2 

2.39 

12 

70.0 

20.58 

.460 

10.00 

538.8 

5.12 

89.8 

114.7 

2.36 

« 

55.0 

16.18 

.370 

9.75 

432.0 

5.17 

72.0 

81.1 

2.24 

10 

44.0 

12.95 

.310 

9.00 

244.2 

4.34 

48.8 

57.3 

2.10 

9 

38.0 

11.22 

.300 

8.50 

170.9 

3.90 

38.0 

44.1 

1.98 

8 

32.5 

9.54 

.290 

8.00 

114.4 

3.46 

28.6 

32.9 

1.86 

TABLES 


xxin 


TABLE  VII 
PROPERTIES  OF  BETHLEHEM  I-BEAMS 


DEPTH 

OF 

BEAM 

WEIGHT 

PER 

FOOT 

AREA 

OF 

SECTION 

THICK- 
NESS OF 
WEB 

WIDTH 

OF 

FLANGE 

NEUTRAL  Axis  PERPEN- 
DICULAR TO  WEB  AT 
CENTER 

NEUTRAL  Axis 
COINCIDENT  WITH 
CENTER  LINE 
OF  WEB 

Inches 

Pounds 

Square 
Inches 

Inches 

Inches 

Moment 
of 
Inertia 

Radius 
of 
Gyration 

Section 
Modulus 

Moment 
of 
Inertia 

Radius 
of 
Gyration 

30 

120.0 

35.30 

.540 

10.500 

5239.6 

12.18 

349.3 

165.0 

2.16 

28 

105.0 

30.88 

.500 

10.000 

4014.1 

11.40 

286.7 

131.5 

2.06 

26 

90.0 

26.49 

.460 

9.500 

2977.2 

10.60 

229.0 

101.2 

1.95 

24 

84.0 

24.80 

.460 

9.250 

2381.9 

9.80 

198.5 

91.1 

1.92 

» 

83.0 

24.59 

.520 

9.130 

2240.9 

9.55 

186.7 

78.0 

1.78 

u 

73.0 

21.47 

.390 

9.000 

2091.0 

9.87 

174.3 

74.4 

1.86 

20 

82.0 

24.17 

.570 

8.890 

1559.8 

8.03 

156.0 

79.9 

1.82 

u 

72.0 

21.37 

.430 

8.750 

1466.5 

8.28 

146.7 

75.9 

1.88 

u 

69.0 

20.26 

.520 

8.145 

1268.9 

7.91 

126.9 

51.2 

1.59 

(( 

64.0 

18.86 

.450 

8.075 

1222.1 

8.05 

122.2 

49.8 

1.62 

(( 

59.0 

17.36 

.375 

8.000 

1172.2 

8.22 

117.2 

48.3 

1.66 

18 

59.0 

17.40 

.495 

7.675 

883.3 

7.12 

98.1 

39.1 

1.50 

(( 

54.0 

15.87 

.410 

7.590 

842.0 

7.28 

93.6 

37.7 

1.54 

u 

52.0 

15.24 

.375 

7.555 

825.0 

7.36 

91.7 

37.1 

1.56 

(( 

48.5 

14.25 

.320 

7.500 

798.3 

7.48 

88.7 

36.2 

1.59 

15 

71.0 

20.95 

.520 

7.500 

796.2 

6.16 

106.2 

61.3 

1.71 

u 

64.0 

18.81 

.605 

7.195 

664.9 

5.95 

88.6 

41.9 

1.49 

u 

54.0 

15.88 

.410 

7.000 

610.0 

6.20 

81.3 

38.3 

1.55 

u 

46.0 

13.52 

.440 

6.810 

484.8 

5.99 

64.6 

25.2 

1.36 

u 

41.0 

12.02 

.340 

6.710 

456.7 

6.16 

60.9 

24.0 

1.41 

u 

38.0 

11.27 

.290 

6.660 

442.6 

6.27 

59.0 

23.4 

1.44 

12 

36.0 

10.61 

.310 

6.300 

269.2 

5.04 

44.9 

21.3 

1.42 

« 

32.0 

9.44 

.335 

6.205 

228.5 

4.92 

38.1 

16.0 

1.30 

it 

28.5 

8.42 

.250 

6.120 

216.2 

5.07 

36.0 

15.3 

1.35 

10 

28.5 

8.34 

.390 

5.990 

134.6 

4.02 

26.9 

12.1 

1.21 

u 

23.5 

6.94 

.250 

5.850 

122.9 

4.21 

24.6 

11.2 

1.27 

9 

24.0 

7.04 

.365 

5.555 

92.1 

3.62 

20.5 

8.8 

1.12 

u 

20.0 

6.01 

.250 

5.440 

85.1 

3.76 

18.9 

8.2 

1.17 

8 

19.5 

5.78 

.325 

5.325 

60.6 

3.24 

15.1 

6.7 

1.08 

" 

17.5 

5.18 

.250 

5.250 

57.4 

3.33 

14.3 

6.4 

1.11 

XXIV 


RESISTANCE  OF  MATERIALS 


TABLE  VIII 

MOMENTS  OF  INERTIA  AND  SECTION  MODULI  :   RECTANGULAR 
CROSS  SECTION 


2 

HEIGHT 
h 

MOMENT 

OF 

INERTIA 

Mi* 
~  12 

SECTION 
MODULUS 
«,     bh* 
6 

1  BREADTH 
b 

HEIGHT 
h 

MOMENT 

OF 

INERTIA 

,     bh* 
12 

SECTION 
MODULUS 

,.» 

BREADTH 
b 

HEIGHT 
h 

MOMENT 

OF 

INERTIA 
/=lf 

SECTION 

MODULI'S 

s=M! 

1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 

.0833 
.66 
2.25 
5.33 
10.42 
18 
28.58 
42.66 
60.75 
83.33 
110.92 
144 

.166 
.66 
1.5 
2.66 
4.16 
6 
8.16 
10.66 
13.5 
16.66 
20.16 
24 

4 

4 
5 
6 
7 
8 
9 
10 
11 
12 

21.33 
41.66 
72 
114.33 
170.66 
243 
333.33 
443.66 
576 

10.66 
16.66 
24 
32.66 
42.66 
54 
66.66 
80.66 
96 

8 
9 

8 
9 
10 
11 
12 
13 
14 
15 
16 

341.33 
486 
666.66 
887.33 
1152 
1464.66 
1829.33 
2250 
2730.66 

85.33 
108 
133.33 
161.33 
192 
225.33 
261.33 
300 
341.33 

5 

5 
6 
7 
8 
9 
10 
11 
12 

52.08 
90 
142.92 
213.33 
303.75 
416.66 
554.58 
720 

20.83 
30 
40.83 
53.33 
67.5 
83.33 
100.83 
120 

9 
10 
11 
12 
13 
14 
15 
16 
17 
18 

546.75 
750 

998.25 
1296 
1647.75 
2058 
2531.25 
3072 
3684.75 
4374 

121.5 
150 
181.5 
216 
253.5 
294 
337.5 
384 
433.5 
486 

2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 

1.33 
4.5 

10.66 
20.83 
36 
57.16 
85.33 
121.5 
166.66 
221.85 
288 

1.33 
3 
5.33 

8.33 
12 
16.33 
21.33 
27 
33.33 
40.33 
48 

6 

6 
7 
8 
9 
10 
11 
12 

108 
171.5 
256 
364.5 
500 
665.5 
864 

36 
49 
64 
81 
100 
121 
144 

10 

10 
11 
12 
13 
14 
15 
16 
17 
18 
19 
20 

833.33 
1109.16 
1440 
1830.83 
2286.66 
2810 
3413.33 
4094.17 
4860 
5715.83 
6666.66 

166.66 
201.66 
240 
281.66 
326.66 
375 
426.66 
481.66 
540 
601.66 
666.66 

3 

3 
4 
5 

6 
7 
8 
9 
10 
11 
12 

6.75 
16 
31.25 
54 
85.75 
128 
182.25 
250 
332.75 
432 

4.5 
8 
12.5 
18 
24.5 
32 
40.5 
50 
60.5 
72 

7 

7 
8 
9 
10 
11 
12 
13 
14 

200.08 
298.66 
425.25 
583.33 
776.42 
1008 
1281.58 
1600.66 

57.16 
74.66 
94.5 
116.66 
141.16 
168 
197.16 
228.66 

TABLES 


XXV 


TABLE  IX 

MOMENTS  OF  INERTIA  AND  SECTION  MODULI  :  CIRCULAR 
CROSS  SECTION 


g  S5 

3S 

s« 

MOMENT 

OF 

INERTIA 

SECTION 
MODULUS 

DIAM- 
ETER 

MOMENT 

OF 

INERTIA 

SECTION 
MODULUS 

DIAM- 
ETER 

MOMENT 

OF 

INERTIA 

SECTION 
MODULUS 

1 

2 
3 
4 
5 
6 
7 
8 
9 
10 

.0491 
.7854 
3.976 
12.57 
30.68 
63.62 
117.9 
201.1 
322.1 
490.9 

.0982 
.7854 
2.651 
6.283 
12.27 
21.21 
33.67 
50.27 
71.57 
98.17 

35 
36 
37 
38 
39 
40 

73,662 
82,448 
91,998 
102,354 
113,561 
125,664 

4,209 

4,580 
4,973 

5,387 
5,824 
6,283 

69 
70 

1,112,660 
1,178,588 

32,251 
33,674 

71 
72 
73 
74 
75 
76 
77 
78 
79 
80 

1,247,393 
1,319^67 
,393,995 
,471,963 
,553,156 
,637,662 
,725,571 
1,816,972 
1,911,967 
2,010,619 

35,138 
36,644 
38,192 
39,783 
41,417 
43,096 
44,820 
46,589 
48,404 
50,265 

41 
42 
43 
44 
45 
46 
47 
48 
49 
50 

138,709 
152,745 
167,820 
183,984 
201,289 
219,787 
239,531 
260,576 
282,979 
306,796 

6,766 
7,274 
7,806 
8,363 
8,946 
9,556 
10,193 
10,857 
11,550 
12,270 

11 
12 
13 
14 
15 
16 
17 
18 
19 
20 

718.7 
1,018 
1,402 
1,886 
2,485 
3,217 
4,100 
5,153 
6,397 
7,854 

130.7 
169.6 
215.7 
269.4 
331.3 
402.1 
482.3 
572.6 
673.4 
785.4 

81 

82 
83 
84 
85 
86 
87 
88 
89 
90 

2,113,051 

2,219,347 
2,329,605 
2,443,920 
2,562,392 
2,685,120 
2,812,205 
2,943,748 
3,079,853 
3,220,623 

52,174 
54,130 
56,135 
58,189 
60,292 
62,445 
64,648 
66,903 
69,210 
71,569 

51 

52 
53 
54 
55 
56 
57 
58 
59 
60 

332,086 
358,908 
387,323 
417,393 
449,180 
482,750 
518,166 
555,497 
594,810 
636,172 

13,023 
13,804 
14,616 
15,459 
16,334 
17,241 
18,181 
19,155 
20,163 
21,206 

21 
22 
23 
24 

25 
26 

27 
28 
29 
30 

9,547 
11,499 
13,737 
16,286 
19,175 
22,432 
26,087 
30,172 
34,719 
39,761 

909.2 
1,045 
1,194 
1,357 
1,534 
1,726 
1,932 
2,155 
2,394 
2,651 

91 
92 
93 
94 
95 
96 
97 
98 
99 
100 

3,366,165 
3,516,586 
3,671,992 
3,832,492 
3,998,198 
4,169,220 
4,345,671 
4,527,664 
4,715,315 
4,908,727 

73,982 
76,448 
78,968 
81,542 
84,173 
86,859 
89,601 
92,401 
95,259 
98,175 

61 
62 
63 
64 
65 
66 
67 
68 

679,651 
725,332 
773,272 
823,550 
876,240 
931,420 
989,166 
1,049,556 

22,284 
23,398 
24,548 
25,736 
26,961 
28,225 
29,527 
30,869 

31 
32 
33 
34 

45,333 
51,472 

58,214 
65,597 

2,925 
3,217 
3,528 
3,859 

TABLE  X 
FOUR-PLACE  LOGARITHMS  OF  NUMBERS 


1 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

0 

0000 

0000 

3010 

4771 

6021 

6990 

7782 

8451 

9031 

9542 

1 

0000 

0414 

0792 

1139 

1461 

1761 

2041 

2304 

2553 

2788 

2 

3010 

3222 

3424 

3617 

3802 

3979 

4150 

4314 

4472 

4624 

3 

4771 

4914 

5051 

5185 

5315 

5441 

5563 

5682 

5798 

5911 

4 

6021 

6128 

6232 

6335 

6435 

6532 

6628 

6721 

6812 

6902 

5 

6990 

7076 

7160 

7243 

7324 

7404 

7482 

7559 

7634 

7709 

6 

7782 

7853 

7924 

7993 

8062 

8129 

8195 

8261 

8325 

8388 

7 

8451 

8513 

8573 

8633 

8692 

8751 

8808 

8865 

8921 

8976 

8 

9031 

9085 

9138 

9191 

9243 

9294 

9345 

9395 

9445 

9494 

9 

9542 

9590 

9638 

9685 

9731 

9777 

9823 

9868 

9912 

9956 

10 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

0374 

11 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

0755 

12 

0792 

0828 

0864 

0899 

0934 

0969 

1004 

1038 

1072 

1106 

13 

1139 

1173 

1206 

1239 

1271 

1303 

1335 

1367 

1399 

1430 

14 

1461 

1492 

1523 

1553 

1584 

1614 

1644 

1673 

1703 

1732 

15 

1761 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

16 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

17 

2304 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

18 

2553 

2577 

2601 

2625 

2648 

2672 

2695 

2718 

2742 

2765 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

22 

3424 

3444 

3464 

3483 

3502 

3522 

3541 

3560 

3579 

3598 

23 

3617 

3636 

3655 

3674 

3692 

3711 

3729 

3747 

3766 

3784 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3909 

3927 

3945 

3962 

25 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133 

26 

4150 

4166 

4183 

4200 

4216 

4232 

4249 

4265 

4281 

4298 

27 

4314 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4456 

28 

4472 

4487 

4502 

4518 

4533 

4548 

4564 

4579 

4594 

4609 

29 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

5011 

5024 

5038 

32 

5051 

5065 

5079 

5092 

5105 

5119 

5132 

5145 

5159 

5172 

33 

5185 

5198 

5211 

5224 

5237 

5250 

5263 

5276 

5289 

5302 

34 

5315 

5328 

5340 

5353 

5366 

5378 

5391 

5403 

5416 

5428 

35 

5441 

5453 

5465 

5478 

5490 

5502 

5514 

5527 

5539 

5551 

36 

5563 

5575 

5587 

5599 

5611 

5623 

5635 

5647 

5658 

5670 

37 

5682 

5694 

5705 

5717 

5729 

5740 

5752 

5763 

5775 

5786 

38 

5798 

5809 

5821 

5832 

5843 

5855 

5866 

5877 

5888 

5899 

39 

5911 

5922 

5933 

5944 

5955 

5966 

5977 

5988 

5999 

6010 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

41 

6128 

6138 

6149 

6160 

6170 

6180 

6191 

6201 

6212 

6222 

42 

6232 

6243 

6253 

6263 

6274 

6284 

6294 

6304 

6314 

6325 

43 

6335 

6345 

6355 

6365 

6375 

6385 

6395 

6405 

6415 

6425 

44 

6435 

6444 

6454 

6464 

6474 

6484 

6493 

6503 

6513 

6522 

45 

6532 

6542 

6551 

6561 

6571 

6580 

6590 

6599 

6609 

6618 

46 

6628 

6637 

6646 

6656 

6665 

6675 

6684 

6693 

6702 

6712 

47 

6721 

6730 

6739 

6749 

6758 

6767 

6776 

6785 

6794 

6803 

48 

6812 

6821 

6830 

6839 

6848 

6857 

6866 

6875 

6884 

6893 

49 

6902 

6911 

6920 

6928 

6937 

6946 

6955 

6964 

6972 

6981 

50 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

xxvi 


TABLES 


XXVll 


50 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

50 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

51 

7076 

7084 

7093 

7101 

7110 

7118 

7126 

7135 

7143 

7152 

52 

7160 

7168 

7177 

7185 

7193 

7202 

7210 

7218 

7226 

7235 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7300 

7308 

7316 

54 

7324 

7332 

7340 

7348 

7356 

7364 

7372 

7380 

7388 

7396 

55 

7404 

7412 

7419 

7427 

7435 

7443 

7451 

7459 

7466 

7474 

56 

7482 

7490 

7497 

7505 

7513 

7520 

7528 

7536 

7543 

7551 

57 

7559 

7566 

7574 

7582 

7589 

7597 

7604 

7612 

7619 

7627 

58 

7634 

7642 

7649 

7657 

7664 

7672 

7679 

7686 

7694 

7701 

59 

7709 

7716 

7723 

7731 

7738 

7745 

7752 

7760 

7767 

7774 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

61 

7853 

7860 

7868 

7875 

7882 

7889 

7896 

7903 

7910 

7917 

62 

7924 

7931 

7938 

7945 

7952 

7959 

7966 

7973 

7980 

7987 

63 

7993 

8000 

8007 

8014 

8021 

8028 

8035 

8041 

8048 

8055 

64 

8062 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

66 

8195 

8202 

8209 

8215 

8222 

8228 

8235 

8241 

8248 

8254 

67 

8261 

8267 

8274 

8280 

8287 

8293 

8299 

8306 

8312 

8319 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

8370 

8376 

8382 

69 

8388 

8395 

8401 

8407 

8414 

8420 

8426 

8432 

8439 

8445 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

71 

8513 

8519 

8525 

8531 

8537 

8543 

8549 

8555 

8561 

8567 

72  ' 

8573 

8579 

8585 

8591 

8597 

8603 

8609 

8615 

8621 

8627 

73 

8633 

8639 

8645 

8651 

8657 

8663 

8669 

8675 

8681 

8686 

74 

8692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

75 

8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

76 

8808 

8814 

8820 

8825 

8831 

8837 

8842 

8848 

8854 

8859 

77 

8865 

8871 

8876 

8882 

8887 

8893 

8899 

8904 

8910 

8915 

78 

8921 

8927 

8932 

8938 

8943 

8949 

8954 

8960 

8965 

8971 

79 

8976 

8982 

8987 

8993 

8998 

9004 

9009 

9015 

9020 

9025 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

81 

9085 

9090 

9096 

9101 

9106 

9112 

9117 

9122 

9128 

9133 

82 

9138 

9143 

9149 

9154 

9159 

9165 

9170 

9175 

9180 

9186 

83 

9191 

9196 

9201 

9206 

9212 

9217 

9222 

9227 

9232 

9238 

84 

9243 

9248 

9253 

9258 

9263 

9269 

9274 

9279 

9284 

9289 

85 

9294 

9299 

9304 

9309 

9315 

9320 

9325 

9330 

9335 

9340 

86 

9345 

9350 

9355 

9360 

9365 

9370 

9375 

9380 

9385 

9390 

87 

9395 

9400 

9405 

9410 

9415 

9420 

9425 

9430 

9435 

9440 

88 

9445 

9450 

9455 

9460 

9465 

9469 

9474 

9479 

9484 

9489 

89 

9494 

9499 

9504 

9509 

9513 

9518 

9523 

9528 

9533 

9538 

90 

9542 

9547 

9552 

9557 

9562 

9566 

9571 

9576 

9581 

9586 

91 

9590 

9595 

9600 

9605 

9609 

9614 

9619 

9624 

9628 

9633 

92 

9638 

9643 

9647 

9652 

9657 

9661 

9666 

9671 

9675 

9680 

93 

9685 

9689 

9694 

9699 

9703 

9708 

9713 

9717 

9722 

9727 

94 

9731 

9736 

9741 

9745 

9750 

9754 

9759 

9763 

9768 

9773 

95 

9777 

9782 

9786 

9791 

9795 

9800 

9805 

9809 

9814 

9818 

96 

9823 

9827 

9832 

9836 

9841 

9845 

9850 

9854 

9859 

9863 

97 

9868 

9872 

9877 

9881 

9886 

9890 

9894 

9899 

9903 

9908 

98 

9912 

9917 

9921 

9926 

9930 

9934 

9939 

9943 

9948 

9952 

99 

9956 

9961 

9965 

9969 

9974 

9978 

9983 

9987 

9991 

9996 

100 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

XXV111 


RESISTANCE  OF  MATERIALS 


TABLE  XI 

CONVERSION  OF  LOGARITHMS 

KEDUCTION  OF  COMMON  LOGARITHMS  TO  NATURAL  LOGARITHMS 

Rule  for  using  Table.  Divide  the  given  common  logarithm  into  periods  of  two 
digits  and  take  from  the  table  the  corresponding  numbers,  having  regard  to  their 
value  as  decimals.  The  sum  will  be  the  required  natural  logarithm. 

Example.  Find  the  natural  logarithm  corresponding  to  the  common  logarithm 
.497149. 

COMMON  LOGARITHMS  NATURAL  LOGARITHMS 


.49 
.0071 
.000049 
.497149 


1.1282667 
.016348354 
.00011282667 

1.14472788067 


COM- 
MON 
LOGA- 
RITHM 

NATURAL 
LOGARITHM 

COM- 
MON 
LOGA- 
RITHM 

NATURAL 
LOGARITHM 

COM- 
MON 
LOGA- 
RITHM 

NATURAL 
LOGARITHM 

COM- 
MON 
LOGA- 
RITHM 

NATURAL 
LOGARITHM 

1 

2.30259 

26 

59.86721 

51 

117.43184 

76 

174.99647 

2 

4.60517 

27 

62.16980 

52 

119.73442 

77 

177.29905 

3 

6.90776 

28 

64.47238 

53 

122.03701 

78 

179.60164 

4 

9.21034 

29 

66.77497 

54 

124.33959 

79 

181.90422 

5 

11.51293 

30 

69.07755 

55 

126.64218 

80 

184.20681 

6 

13.81551 

31 

71.38014 

56 

128.94477 

81 

186.50939 

7 

16.11810 

32 

73.68272 

57 

131.24735 

82 

188.81198 

8 

18.42068 

33 

75.98531 

58 

133.54994 

83 

191.11456 

9 

20.73327 

34 

78.28789 

59 

135.85252 

84 

193.41715 

10 

23.02585 

35 

80.59048 

60 

138.15511 

85 

195.71973 

11 

25.32844 

36 

82.89306 

61 

140.45769 

86 

198.02232 

12 

27.63102 

37 

85.19565 

62 

142.76028 

87 

200.32490 

13 

29.93361 

38 

87.49823 

63 

145.06286 

88 

202.62749 

14 

32.23619 

39 

89.80082 

64 

147.36545 

89 

204.93007 

15 

34.53878 

40 

92.10340 

65 

149.66803 

90 

207.23266 

16 

36.84136 

41 

94.40599 

66 

151.97062 

91 

209.53524 

17 

39.14395 

42 

96.70857 

67 

154.27320 

92 

211.83783 

18 

41.44653 

43 

99.01116 

68 

156.57579 

93 

214.14041 

19 

43.74912 

44 

101.31374 

69 

158.87837 

94 

216.44300 

20 

46.05170 

45 

103.61633 

70 

161.18096 

95 

218.74558 

21 

48.35429 

46 

105.91891 

71 

163.48354 

96 

221.04817 

22 

50.65687 

47 

108.22150 

72 

165.78613 

97 

223.35075 

23 

52.95946 

48 

'  110.52408 

73 

168.08871 

98 

225.65334 

24 

55.26204 

49 

112.82667 

74 

170.39130 

99 

227.95592 

25 

57.56463 

50 

115.12925 

75 

172.69388 

100 

230.25851 

TABLES 


xxix 


TABLE  XII 
FUNCTIONS  OF  ANGLES 


ANGLE 

SIN 

TAN 

SEC 

COSEC 

COT 

Cos 

0 

0. 

0. 

.0 

CO 

GO 

1. 

90 

1 

0.0175 

0.0175 

1.0001 

57.299 

57.290 

0.9998 

89 

2 

.0349 

.0349 

1.0006 

28.654 

28.636 

.9994 

88 

3 

.0523 

.0524 

1.0014 

19.107 

19.081 

.9986 

87 

4 

.0698 

.0699 

.0024 

14.336 

14.301 

.9976 

86 

5 

.0872 

.0875 

1.0038 

11.474 

11.430 

.9962 

85 

6 

0.1045 

0.1051 

1.0055 

9.5668 

9.5144 

0.9945 

84 

7 

.1219 

.1228 

.0075 

8.2055 

8.1443 

.9925 

83 

8 

.1392 

.1405 

.0098 

7.1853 

7.1154 

.9903 

82 

9 

.1564 

.1584 

.0125 

6.3925 

6.3138 

.9877 

81 

10 

.1736 

.1763 

1.0154 

5.7588 

5.6713 

.9848 

80 

11 

0.1908 

0.1944 

1.0187 

5.2408 

5.1446 

0.9816 

79 

12 

.2079 

.2126 

1.0223 

4.8097 

4.7046 

.9781 

78 

13 

.2250 

.2309 

1.0263 

4.4454 

4.3315 

.9744 

77 

14 

.2419 

.2493 

1.0306 

4.1336 

4.0108 

.9703 

76 

15 

.2588 

.2679 

"1.0353 

3.8637 

3.7321 

.9659 

75 

16 

0.2756 

0.2867 

1.0403 

3.6280 

3.4874 

0.9613 

74 

17 

.2924 

.3057 

1.0457 

3.4203 

3.2709 

.9563 

73 

18 

.3090 

.3249 

1.0515 

3.2361 

3.0777 

.9511 

72 

19 

.3256 

.3443 

1.0576 

3.0716 

2.9042 

.9455 

71 

20 

.3420 

.3640 

1.0642 

2.9238 

2.7475 

.9397 

70 

21 

0.3584 

0.3839 

1.0712 

2.7904 

2.6051 

0.9336 

69 

22 

.3746 

.4040 

1.0785 

2.6695 

2.4751 

.9272 

68 

23 

.3907 

.4245 

1.0864 

2.5593 

2.3559 

.9205 

67 

24 

.4067 

.4452 

.0946 

2.4586 

2.2460 

.9135 

66 

25 

.4226 

.4663 

.1034 

2.3662 

2.1445 

.9063 

65 

26 

0.4384 

0.4877 

.1126 

2.2812 

2.0503 

0.8988 

64 

27 

.4540 

.5095 

.1223 

2.2027 

1.9626 

.8910 

63 

28 

.4695 

.5317 

.1326 

2.1301 

1.8807 

.8829 

62 

29 

.4848 

.5543 

.1434 

2.0627 

1.8040 

.8746 

61 

30 

.5000 

.5774 

.1547 

2.0000 

1.7321 

.8660 

60 

31 

0.5150 

0.6009 

.1666 

1.9416 

1.6643 

0.8572 

59 

32 

.5299 

.6249 

.1792 

1.8871 

1.6003 

.8480 

58 

33 

.5446 

.6494 

1.1924 

1.8361 

1.5399 

.8387 

57 

34 

.5592 

.6745 

1.2062 

1.7883 

.4826 

.8290 

56 

35 

.5736 

.7002 

1.2208 

1.7435 

.4281 

.8192 

55 

36 

0.5878 

0.7265 

1.2361 

1.7013 

.3764 

0.8090 

54 

37 

.6018 

.7536 

1.2521 

1.6616 

.3270 

.7986 

53 

38 

.6157 

.7813 

1.2690 

1.6243 

.2799 

.7880 

52 

39 

.6293 

.8098 

1.2868 

1.5890 

.2349 

.7771 

51 

40 

.6428 

.8391 

1.3054 

1.5557 

.1918 

.7660 

50 

41 

0.6561 

0.8693 

1.3250 

1.5243 

.1504 

0.7547 

49 

42 

.6691 

.9004 

1.3456 

1.4945 

.1106 

.7431 

48 

43 

.6820 

.9325 

1.3673 

1.4663 

1.0724 

.7314 

47 

44 

.6947 

.9657 

1.3902 

1.4396 

1.0355 

.7193 

46 

45 

.7071 

1. 

1.4142 

1.4142 

1. 

.7071 

45 

Cos 

COT 

COSEC 

SEC 

TAN 

SIN 

ANGLE 

XXX 


RESISTANCE  OF  MATERIALS 


TABLE  XIII 
BENDING  MOMENT  AND  SHEAR  DIAGRAMS 


Jfa  =  Jfc  =  0. 


l/t 


MOMENT 


Bt  =  B2  =  P. 
Jfa  =  Jfc  =  0. 

Mb  =  Jf E  =  Pa 


4SEI 


384  J£Z 


Mmx=Rld-P(d-a). 


TABLES 


XXXI 


i      : 


i ^ 


I     '    I2  MOMENT 


SUEAR 

3  Pd2 


3  El  P 


Pi 


.R  =  P. 
MA=-PL 

p/3 


u 


=  wl  +  P. 


D  = 


XXX11 


RESISTANCE  OF  MATERIALS 


MB  =  &  PL 


"An 


Wl 


MOMENT 


SHEAR\    \OjL 


wl* 


384^1 


MOMENTS'      | 

i r> 


SHEAR 


Mb=-Pd. 
Mc=-Pd. 


Ik    i4\i  fi' 


TABLES 


xxxiu 


TABLE  XIV 

MENSURATION 
CIRCULAR  MEASURE 

Circumference  of  circle  =  diameter  x  3.1416. 

Diameter  of  circle  =  circumference  x  0.3183. 

Side  of  square  of  same  periphery  as  circle  =  diameter  x  0.7854. 

Diameter  of  circle  of  same  periphery  as  square  =  side  x  1.2732. 

Side  of  inscribed  square  =  diameter  of  circle  x  0.7071. 

Length  of  arc  =  number  of  degrees  x  diameter  x  0.008727. 

Circumference  of  circle  whose  diameter  is  1  =  TT. 


v  =  r  —  \  r2 ,  or  very  nearly  = 


r  =  radius 
c  =  chord 
v  —  versine 
o  =  ordinate 


or  very  nearly  = 


NUMBER 
?r  =  3.14159265 
VTT  =  1.772454. 
7T2  =  9.869604 

-  =  0.318310. 

TT 

-L  =  0.101321. 
—  =  0.564190 


COMMON  LOGARITHM 
0.4971499 
0.24857494 
0.99429995 

9.50285013  -  10 
9.00570025  -  10 
9.75142506-  10 


1  radian  =  angle  subtended  by  circular  arc  equal  in  length  to  the  radius  of  the 
circle  ;  TT  radians  =  180  degrees  ; 


1  radian  =  ( -  57.29577951°. 

y  TT  i 

Segment  of  circle  —  area  of  sector  less  triangle  ;   also  for  flat  segments  very 


4  „      /  rz 

nearly  =  —  -*/  0. 388  v2  +  - . 
o    \  4 


Side  of  square  of  same  area  as  circle  =  diameter  x  0.8862  ;  also  =  circumference 
x  0.2821. 

Diameter  of  circle  of  same  area  as  square  =  side  x  1.1284. 

Area  of  parabola  =  base  x  §  height. 

Area  of  ellipse  =  long  diameter  x- short  diameter  x  0.7854. 

Area  of  regular  polygon  =  sum  of  sides  x  half  perpendicular  distance  from  center 
to  sides. 


XXXIV 


KESISTANCE  OF  MATERIALS 


TABLE  XV 

FRACTIONAL  AND  DECIMAL  EQUIVALENTS 
FRACTIONS  OF  A  LINEAL  INCH  IN  DECIMALS 


FRAC- 
TIONS 

DECIMALS  OF 
AN  INCH 

FRAC- 
TIONS 

DECIMALS  OF 
AN  INCH 

FRAC- 
TIONS 

DECIMALS  OF 
AN  INCH 

FRAC- 
TIONS 

DECIMALS  OF 
AN  INCH 

A 

0.015625 

u 

0.265625 

fl 

0.515635 

If 

0.765625 

A 

0.03125 

A 

0.28125 

II 

0.53125 

M 

0.78125 

A' 

0.04687 

H 

0.296875 

fl 

0.546875 

tt 

0.796875 

TV 

0.0625 

0.3125 

& 

0.5625 

H 

0.8125 

A 

0.078125 

ti 

0.328125 

fl- 

0.578125 

fl 

0.828125  • 

A 

0.09375 

si 

0.34375 

it 

0.59375 

II 

0.84375 

& 

0.109375 

|| 

0.359375 

if 

0.609375 

II 

0.859375 

4 

0.125 

I 

0.375 

i 

0.625 

i 

0.875 

A 

0.140625 

II 

0.390625 

ft 

0.640625 

fl 

0.890625 

A 

0.15625 

41 

0.40625 

'  Si 

0.65625 

il 

0.90625 

¥ 

0.171875 

H 

0.421875 

II 

0.671875 

£a 

6  4 

0.921875 

0.1875 

A 

0.4375 

w 

0.6875 

if 

0.9375 

jt 

0.203125 

H 

0.453125 

fl 

0.703125 

If 

0.953125 

A 

0.21875 

1  5 

0.46875 

11 

0.71875 

0.96875 

II 

0.234375 

II 

0.484375 

« 

0.734375 

ft 

0.984375 

i 

0.25 

i 

0.5 

i 

0.75 

1 

1.000 

LINEAL  INCHES  IN  DECIMAL  FRACTIONS  OF  A  LINEAL  FOOT 


LINEAL 
INCHES 

LINEAL  FOOT 

LINEAL 
INCHES 

LINEAL  FOOT 

LINEAL 
INCHES 

LINEAL  FOOT 

A 

0.001302083 
0.00260416 

2 

0.15625 
0.1666 

7* 

0.5625 
0.5833 

iV 

0.0052083 

2J 

0.177083 

7? 

0.60416 

^ 

0.010416 

21 

0.1875 

7i 

0.625 

T\ 

0.015625 

2| 

0.197916 

7| 

0.64583 

-1 

0.02083 

2i 

0.2088 

8 

0.66667 

t 

0.0260416 
0.03125 
0.0364583 

2| 

2I 

0.21875 
0.22916 
0.239583 

4 
8| 

0.6875 
0.7083 
0.72916 

£ 

0.0416 

3 

0.25 

9 

0.75 

T\ 

0.046875 

31 

0.27083 

91 

0.77083 

f 

V 

it 

0.052083 
0.0572916 
0.0625 
0.0677083 

3| 
4 

0.2916 
0.3125 
0.33333 
0.35416 

4 

9| 
10 
101 

0.7916 
0.8125 
0.83333 
0.85416 

0.072916 

4i 

0.375 

10J 

0.875 

•' 

0.078125 
0.0833 

4| 
5 

0.39583 
0.4166 

11 

0.89583 
0.9166 

h-  '  1—  '  1—  '  1—  '  1—  » 
JtCnNHolWiUMOcH 

0.09375 
0.10416 
0.114583 
0.125 
0.135416 

6| 

6 

0.4375 
0.4583 
0.47916 
0.5 
0.52083 

11J 
114 
111 

12 

0.9375 
0.9583 
0.97916 
1.000 

I 

0.14583 

si 

0.5416 

4 

TABLES 


xxxv 


TABLE  XVI 
WEIGHTS  OF  VARIOUS  SUBSTANCES 


MATERIAL 

WEIGHT  IN 

LB./FT.3 

Pressed  brick 

150 

Brick  and 
brickwork 

Common  hard  brick 
Soft  inferior  brick 
Good  pressed-brick  masonry 

125 
100 
140 

Ordinary  brickwork 

125 

Gneiss,  solid 

168 

Gneiss,  loose  piles 

96 

Granite 

170 

Limestone  and  marble 

165 

Limestone  and  marble,  loose,  broken 

96 

Sandstone,  solid 

151 

Stone  and 

Sandstone,  quarried  and  piled 

86 

masonry 

Shale 

162 

Slate 

175 

Granite  or  limestone  masonry,  well  dressed 

165 

Granite^or  limestone  masonry,  mortar  rubble 

154 

Granite  or  limestone  masonry,  well-scabbled 

dry  rubble 

138 

Sandstone  masonry,  well  dressed 

144 

Common  loam,  dry,  loose 

76 

Common  loam,  moderately  rammed 

95 

Earth,-  sand, 
and  gravel 

Soft  flowing  mud 
Dry  hard  mud 
Gravel  or  sand,  dry,  loose 

110 
80-110 
90-106 

. 

Gravel  or  sand,  well  shaken 

99-117 

Gravel  or  sand,  wet 

120-140 

Aluminium 

162 

Brass,  cast  (copper  and  zinc) 

504 

Brass,  rolled 

524 

Bronze  (copper  8,  tin  1) 

529 

Copper,  cast 

542 

Copper,  rolled 

555 

Metals 

Iron,  cast 

450 

Iron,  wrought 

480 

Lead 

710 

Platinum 

1342 

Steel 

489.6 

Tin,  cast 

459 

Zinc,  spelter 

437.5 

Anthracite,  solid  Pennsylvania 

93 

Anthracite,  broken,  loose  (heaped  bushel  80  Ib.) 

54 

Anthracite,  broken,  shaken 

58 

Coal 

Bituminous,  solid 

84 

Bituminous,  loose  (heaped  bushel  74  Ib.) 

49 

Bituminous,  broken,  shaken 

51-56 

Coke,  loose  (heaped  bushel  40  Ib.) 

26 

XXXVI 


RESISTANCE  OF  MATERIALS 


WEIGHTS  OF  VARIOUS  SUBSTANCES  —  Continued 


MATERIAL 

WEIGHT  IN 

LB./FT.3 

Quicklime,  loose  (struck  bushel  66  Ib.) 

53 

Quicklime,  well  shaken 

75 

American  Louisville,  loose 

50 

Lime 

American  Rosendale,  loose 

56 

and  hydraulic 

American  Cumberland,  loose 

65 

cement 

American  Cumberland,  well  shaken 

85 

English  Portland 

90 

American  Portland,  loose 

88 

American  Portland,  well  shaken 

110 

Ash,  American  white 

38 

Cherry 

42 

Chestnut 

41 

Cypress 

64 

Ebony 

76 

Elm 

35 

Hemlock 

25 

Hickory 

53 

Lignum-vitae 

83 

Locust 

44 

Mahogany,  Spanish 

53 

Dry  wood 

Mahogany,  Honduras 

35 

Maple 

49 

Oak,  live 

59 

Oak,  red  or  black 

32-45 

Oak,  white 

48 

Pine,  white 

•25 

Pine,  yellow  Northern 

34 

Pine,  yellow  Southern 

45 

Poplar 

29 

Sycamore 

37 

Spruce 

25 

Walnut,  black 

38 

TABLES 


XXXVll 


TABLE  XVII 

STRENGTH  OF  ROPES  AND  BELTS 
TENSION  TESTS  OF  STEEL  WIRE  ROPE 


CIRCUMFER- 
ENCE 

in, 

NUMBER 

OF 

STRANDS 

WIRES 

PER 

STRAND 

MEAN 
DIAMETER 
OF  WIRES 

in. 

CORE 

SECTIONAL 
AREA  OF 
WIRE 

in.  2 

TENSILE  STRENGTH 

Total 
Ib. 

Total 
lb./in.2 

1.5 
1.75 

2 
2.125 
2.25 
2.50 
3 
3.50 
4.50 

CO  CO  CO  CO  CO  CO  CO  CO  CO 

18 
18 
18 
18 
18 
18 
18 
18 
18 

.0321 
.0349 
.0420 
.0456 
.0488 
.0544 
.0598 
.0718 
.0980 

Hemp 
Hemp 
Hemp 
Hemp 
Hemp 
Hemp 
Hemp 
Hemp 
Hemp 

.0876 
.1031 
.1499 
.17G6 
.2021 
.2510 
.3024 
.4380 
.8151 

12,898 
15,736 
20,780 
24,430 
30,960 
33,270 
46,370 
65,120 
138,625 

147,236 
153,893 
138,360 
138,383 
148,650 
132,500 
153,340 
148,675 
170,075 

STRENGTH  OF  IRON  WIRE  ROPE* 

(Rope  composed  of  six  strands  and  a  hemp  center,  seven  or  twelve  wires  in  each  strand) 


DIAMETER 
in. 

CIRCUMFERENCE 
in. 

APPROXIMATE  BREAK- 
ING STRENGTH 

CIRCUMFERENCE 
IN  INCHES  OF  NEW 
MANILA  ROPE 

Ib. 

OF  EQUAL  STRENGTH 

1.75 

5.50 

88,000 

11 

1.625 

5.00 

72,000 

10 

1.50 

4.75 

64,000 

9.5 

1.375 

4.25 

52,000 

8.5 

1.25 

4.00 

46,000 

8.0 

1.125 

3.50 

36,000 

6.5 

1.000 

3.00 

26,000 

5.75 

.875 

2.75 

22,000 

5.25 

.750 

2.25 

14,600 

4.75 

.500 

1.50 

6,400 

3.00 

.375 

1.125 

3,600 

2.25 

,250 

.75 

1,620 

1.50 

STRENGTH  OF  CAST  STEEL  WIRE  ROPE* 

(Rope  composed  of  six  strands  and  a  hemp  center,  seven  or  nineteen  wires  in  each  strand) 


DIAMETER 
in. 

CIRCUMFERENCE 
in. 

APPROXIMATE  BREAK- 
ING STRENGTH 

Ib. 

CIRCUMFERENCE 
IN  INCHES  OF  NEW 
MANILA  ROPE 
OF  EQUAL  STRENGTH 

1.25 
2.125 
1.00 
.875 
.750 
.625 
.500 
.375 

4.00 
3.50 
3.00 
2.75 
2.25 
2.00 
2.50 
1.125 

106,000 
82,000 
62,000 
52,000 
35,200 
28,000 
16,200 
9,000 

13 
11 
9 
8.5 
7.0 
6.0 
4.75 
3.75 

*  As  given  by  John  A.  Roebling. 


xxxviii  BESISTANCE  OF  MATERIALS 

TABLE  XVII  (Continued) 

TESTS   OF   MANILA   AND   SISAL   ROPE 
MANILA  ROPE 


SIZE  OF  ROPE 

DIAMETER 

in. 

SECTIONAL, 
AREA 

in.z 

TENSILE  STRENGTH 

TOTAL 
LOAD 

Ib. 

lb./in.2 

Per  Yarn 

Ib. 

6-thread  .    .              ... 

.27 
.30 
.38 
.43 
.49 
.56 
.61 
.62 
.74 
.79 
.78 
.85 
.96 
1.00 
.99 
1.13 
1.19 
1.29 
1.28 
1.39 
1.34 
1.41 
1.59 
1.61 
1.66 
1.76 
2.25 
2.52 
2.83 
3.35 
3.70 

.0567 
.0750 
.114 
.153 
.192 
.259 
.288 
.299 
.41 
.478 
.462 
.557 
.715 
.782 
.746 
.970 
1.07 
1.27 
1.26 
1.46 
1.36 
1.51 
1.88 
1.99 
2.04 
2.35 
3.82 
4.86 
6.22 
8.37 
10.06 

13,360 
14,180 
12,920 
14,250 
11,610 
11,970 
10,800 
11,500 
9,200 
12,900 
11,900 
12,470 
12,810 
13,630 
13,750 
12,470 
12,190 
11,990 
11,610 
10,080 
11,790 
9,890 
10,360 
10,480 
10,740 
9,940 
8,260 
9,400 
8,600 
7,500 
7,300 

126 
118 
123 
145 
125 
148 
130 
128 
114 
148 
138 
136 
153 
146 
151 
144 
132 
134 
132 
117 
130 
113 
121 
134 
128 
118 
112 
125 
118 
108 
102 

750 
1,064 
1,473 
2,180 
2,242 
3,100 
3,120 
3,455 
3,775 
6,207 
5,509 
6,947 
9,160 
10,663 
10,260 
12,093 
13,050 
15,227 
14,640 
14,723 
16,017 
14,943 
19,577 
20,873 
21,903 
23,360 
31,570 
45,647 
54,000 
62,717 
73,910 

9-thread                .    . 

12-thread              .     . 

15-thread              .         .    - 

1  25-in 

1  50-in 

1  625-in 

1.75-in  
2-in  

2.25-in      

2  25-in.              ... 

2  50-in 

2  75-in 

3-in      . 

3-in  

3  25-in      

3  50-in  

3  75-in.     . 

3  75-in. 

4-in. 

4-in 

4.25-in  

4  50-in 

4  50-in  

4  75-in  

5-in.     . 

6-in.     .    . 

7-in.         .                  .    . 

8-in 

9-in 

10-in 

SISAL  ROPE 


SIZE  OF  ROPE 

DIAMETER 
In. 

SECTIONAL 
AREA 

in.2 

TENSILE  STRENGTH 

TOTAL 
LOAD 

Ib. 

Ib./in.a 

Per  Yarn 
Ib. 

.27 
.33 
.39 
.45 
.56 
.63 
.70 
.81 
.95 
1.01 
1.22 

.0567 
.082 
.126 
.129 
.254 
.302 
.395 
.416 
.691 
.780 
1.128 

7,700 
7,300 
7,500 
10,810 
8,100 
7,600 
7,200 
9,500 
8,300 
7,500 
7,200 

72 
67 
79 
93 
99 
96 
97 
94 
101 
104 
102 

432 
605 
944 
1397 
2067 
2315 
2925 
3966 
5733 
5917 
8230 

12-thread     .    .         .    . 

1  25-in  

1  50-in  

1  75-in.     ... 

2-in      .         ... 

2.25-in  

2.75-in  

3-in  

3.50-in  

TABLES 


XXXIX 


TABLE  XVII  (Continued) 
TESTS   OF   RUBBER   BELTING 


DIMENSIONS 

TENSILE  STRENGTH 

in. 

SECTIONAL 

DESCRIPTION 

AREA 

Tliick- 

in.2 

Pound  per 

Length 

Width 

I16SS 

Ib./in.z 

Inch  of 

Width 

2-in.,  4-ply    . 

60.17 

2.02 

.26 

.525 

3276 

851 

6-in.,  4-ply    .     .     . 

60.17 

6.08 

.26 

1.58 

3227 

839 

6-in.,  4-ply    .     .     . 

60.12 

6.13 

.26 

1.59 

3773 

979 

6-in.,  4-ply    .     .     . 

60.17 

6.05 

.26 

1.57 

2739 

711 

12-in.  ,  4-ply    .     .     . 

60.02 

12.08 

.27 

3.26 

3037 

819 

12-in.,  4-ply    .     .     . 

60.14 

12.24 

.26 

3.18 

2987 

776 

2-in.,  6-ply    .     .     . 

60.17 

2.14 

.36 

.770 

3104 

1116 

6-in.,  6-ply    .     .     . 

59.98 

6.26 

.37 

2.32 

2737 

1014 

6-in.,  6-ply    .     .     . 

60.08 

6.27 

.36 

2.26 

3770 

1358 

12-in.,  6-ply    .     .     . 

60.15 

12.04 

.36 

4.33 

3436 

1236 

12-in.,  6-ply    .     .     . 

60.17 

12.16 

.34 

4.13 

3862 

1311 

24-in.,  6-ply    .     .     . 

60.13 

24.11 

.41 

9.89 

2381 

977 

30-in.,  6-ply    .     .     . 

60.04 

30.18 

.40 

12.07 

2808 

1123 

TESTS   OF   LEATHER   BELTING 


DIMENSIONS 

TENSILE  STRENGTH 

in. 

SECTIONAL 

DESCRIPTION  * 

AREA 

Thick- 

in.z 

Pounds  per 

Length 

Width 

lb./in.* 

Inch  of 

Width 

2-in.,  single  .     .     . 

60.00 

1.98 

.20 

.396 

5045 

1091 

6-in.,  single  .     .     . 

60.20 

6.07 

.22 

1.34 

2537 

560 

6-in.,  single  (w) 

60.11 

6.08 

.24 

1.46 

2119 

633 

12-in.,  single  .     .     . 

60.11 

12.05 

.18 

2.17 

3917 

705 

4-in.,  double      .     . 

59.55 

3.98 

.33 

1.31 

4931 

1623 

6-in.,  double      .     . 

60.18 

5.91 

.47 

2.78 

4309 

2027 

6-in.,  double  (w)    . 

59.93 

6.00 

.40 

2.40 

6166 

2066 

12-in.,  double      .     . 

59.90 

11.90 

.39 

4.64 

4090 

1595 

12-in.,  double  (w)     . 

60.06 

11.93 

.36 

4.29 

4424 

1591 

24-in.,  double  (w)    . 

60.00 

23.90 

.47 

11.23 

2760 

1297 

30-in.,  double      .     . 

59.90 

29.95 

.43 

12.88 

2717 

1169 

*  The  letter  w  in  the  table  stands  for  waterproofed. 


RESISTANCE  OF  MATERIALS 

SECTION  I 

STRESS  AND  DEFORMATION 

1.  Elastic  resistance,  or  stress.  The  effect  of  an  external  force 
acting  upon  an  elastic  body  is  to  produce  deformation,  or  change  of 
shape.  For  example,  if  a  bar  is  placed  in  a  testing  machine  and  a 
tensile  load  applied,  it  will  be  found  that  the  length  of  the  bar  is 
increased  and  the  area  of  its  cross  section  diminished.  Similarly,  if 
a  compressive  load  is  applied,  the  length  of  the  bar  is  diminished 
and  the  area  of  its  cross  section  increased. 

All  solid  bodies  offer  more  or  less  resistance  to  the  deformation, 
or  change  of  shape,  produced  by  external  force.  This  internal  resist- 
ance, when  expressed  in  definite  units,  is  called  stress.  A  body  under 
the  action  of  stress  is  said  to  be  strained. 

In  general  the  stress  is  not  the  same  at  all  points  of  a  body,  but 
varies  from  point  to  point.  The  intensity  of  the  stress  at  any  par- 
ticular point  is  therefore  expressed  as  the  force  in  pounds  which 
would  be  exerted  if  the  stress  were  uniform  and  acted  over  an  area 
one  square  inch  in  extent.  That  is  to  say,  whatever  the  actual  extent 
of  the  area  considered,  whether  finite  or  infinitesimal,  the  stress  is 
expressed  in  pounds  per  square  inch  (abbreviated  into  lb./in.2). 

For  example,  suppose  that  a  wire  ^  in.  in  diameter  is  pulled 
with  a  force  of  50  Ib.  Then  for  equilibrium  the  total  stress  acting 
on  any  cross  section  of  the  wire  must  also  be  50  Ib.  But  since  the 
area  of  the  cross  section  is  only  .049  in.2,  the  intensity  of  the  stress 

50 

is  -7777:  >  or  1000  lb./in.2    In  other  words,  if  the  wire  were  1  sq.  in. 
.U4y 

in  cross  section,  the  strain  under  a  load  of  1000  Ib.  would  be 
the  same  as  that  produced  by  a  load  of  50  Ib.  on  a  wire  ^  in. 
in  diameter. 

1 


RESISTANCE  OF  MATERIALS 


Taking  any  plane  section  of  a  body  under  strain,  the  stress  act- 
ing on  this  plane  section  may  in  general  be  resolved,  like  any  force, 
into  two  components,  one  perpendicular  to  the  plane  and  the  other 
lying  in  the  plane.  The  perpendicular,  or  normal,  component  is 
called  direct  stress  and  is  either  tension  or  compression.  The 
tangential  component,  or  that  lying  in  the  plane  of  the  cross 
section,  is  called  shear.  In  what  follows,  the  letter  p  will  always 
be  used  to  denote  normal,  or  direct,  stress,  and  q  to  denote  tan- 
gential stress,  or  shear. 


Tension 


Compression  Buckling 

d 


The  effect  of  a  normal  stress  is  to  produce  extension  or  com- 
pression, that  is,  a  lengthening  or  shortening  of  the  fibers,  thereby 
changing  the  dimensions  of  the  body ;  whereas  shear  tends  to  slide 
any  given  cross  section  over  the  one  adjacent  to  it,  thus  producing 
angular  deformation,  or  change  in  shape,  of  the  body  without 
altering  its  dimensions. 

2.  Varieties  of  strain.  The  nature  of  the  deformation  produced 
by  external  forces  acting  on  an  elastic  body  depends  on  where  and 
how  these  forces  are  applied.  Although  only  two  kinds  of  stress 
can  occur,  —  namely,  normal  stress  (tension  or  compression)  and 


STRESS  AND  DEFOBMATION  3 

shear,  —  these  may  arise  in  various  ways.  In  general,  five  different 
cases  of  strain  may  be  distinguished,  each  of  which  must  be  con- 
sidered separately.  These  are  as  follows : 

1.  If  the  forces  act  along  the  same  line,  toward  or  away  from 
one  another,  the  strain  is  called  compression  or  tension  (Fig.  1,  a). 

2.  If  the  forces  tend  to  slice  off  a  portion  of  the  body  by  sepa- 
rating it  along  a  surface,  the  strain  is  called  shear  (Fig.  1,  ft). 


10          12         14         16 
EXTENSION,  PER  CENT 

FIG.  2 


18        20         22        24 


3.  If  the  forces  act  transverse  to  the  length  of  the  body  (usually 
perpendicular  to  the  long  axis  of  the  piece),  so  as  to  produce  lateral 
deflection,  the  strain  is  called  bending,  or  flexure  (Fig.  1,  e). 

4.  If  one  dimension  of  the  body  is  large  as  compared  with  the 
other  two,  and  the  forces  act  in  the  direction  of  the  long  dimension 
and  toward  one  another,  the  strain  is  called  buckling,  or  column 
flexure  (Fig.  1,  d). 

5.  If  the  forces  exert  a  twist  on  the  body,  the  strain  is  called 
torsion  (Fig.  1,  e). 

Two  or  more  of  these  simple  strains  may  occur  in  combination, 
as  illustrated  in  Fig.  1,  /. 


4  RESISTAHCE  OF  MATERIALS 

3.  Strain  diagram.    In  the  case  of  tension  or  compression  it  is 
easy  to  show  graphically  the  chief  features  of  the  strain.    Thus, 
suppose  that  a  test  bar  is  placed  in  a  testing  machine,  and  that  the 
total  load  on  the  bar  at  any  instant  is  read  on  the  scale  beam  of 
the  machine,  and  its  corresponding  length  in  inches  is  measured 
with  an  extensometer.    Assuming  that  the  stress  is  uniformly  dis- 
tributed over  any  cross  section  of  the  bar,  the  unit  stress  is  obtained 
by  dividing  the  total  load  in  pounds  acting  on  the  bar  by  the  area 
of  its  cross  section  in  square  inches.    That  is, 

.,  total  load  in  pounds 

(1)  p  =  unit  stress  =  —      — - —  — : ~  

area  ot  cross  section  in  square  inches 

Also,  the  total  deformation,  or  change  in  length,  is  divided  by  the 
original  unstrained  length  of  the  bar,  giving  the  unit  deformation 
in  inches  per  inch.  Let  this  be  denoted  by  s ;  that  is,  let 

.,    -,   .          ,.          change  in  length 

(2)  s  =  unit  deformation  =  —  — . 

original  length 

The  unit  deformation  is  therefore  an  abstract  number.  Moreover, 
both  the  unit  stress  and  the  unit  deformation  are  independent  of  the 
actual  dimensions  of  the  test  bar  and  depend  only  on  the  physical 
properties  of  the  material. 

If,  now,  the  unit  stresses  are  plotted  as  ordinates  and  the  corre- 
sponding unit  deformations  as  abscissas,  a  strain  diagram  is  obtained, 
as  shown  in  Fig.  2.  Such  a  diagram  shows  at  a  glance  the  physical 
properties  of  the  material  it  represents,  as  explained  in  what  follows. 

4.  Hooke's  law.    By  inspection  of  the   curves  in  Fig.   2  it  is 
evident  that  the  strain  diagram  for  each  material  has  certain  char- 
acteristic features.    For  instance,  in  the  case  of  wrought  iron  the 
strain  diagram  from  0  to  A  is  a  straight  line ;  this  means  that  for 
points  between  0  and  A  the  stress  is  proportional  to  the  correspond- 
ing deformation.    That  is  to  say,  within  certain  limits  the  ratio  of 
p  to  s  is  constant,  or 

(3)  f  =  *. 

where  the  constant  E  denotes  the  slope  of  the  initial  line.  This 
important  property  is  known  as  Hooke's  law,  and  the  constant  E 
is  called  Young's  modulus  of  elasticity. 


STRESS  AND  DEFORMATION  5 

The  upper  end  A  of  the  initial  line,  or  point  where  the  diagram 
begins  to  curve,  is  called  the  elastic  limit.  The  point  B,  where  the 
deformation  becomes  very  noticeable,  is  called  the  yield  point.  As 
these  two  points  occur  close  together,  no  distinction  is  made  between 
them  in  ordinary  commercial  testing. 

The  maximum  ordinate  to  the  strain  diagram  represents  the 
greatest  unit  stress  preceding  rupture,  and  is  called  the  ultimate 
strength  of  the  material. 

In  the  case  of  shear  let  q  denote  the  unit  shearing  stress  and 
</>  the  corresponding  angular  deformation  expressed  in  circular 
measure.  Then,  by  Hooke's  law, 

(4)  «  =  G, 

where  G  is  a  constant  for  any  given  material,  called  the  modulus 
of  rigidity,  or  shear  modulus.  For  steel  and  wrought  iron  G  =  A  E, 
approximately. 

Average  values  of  E  and  G  for  various  materials  are  given  in 
Table  I. 

5.  Elastic  limit.  It  is  found  by  experiment  that  as  long  as  the 
stress  does  not  pass  the  elastic  limit,  the  deformation  disappears 
when  the  external  forces  are  removed.  If  the  unit  stress  (or,  more 
properly,  the  unit  deformation)  exceeds  the  elastic  limit,  however, 
then  the  deformation  does  not  entirely  disappear  upon  removal  of 
the  load,  but  the  body  retains  a  permanent  set.  At  the  elastic 
limit,  therefore,  the  body  begins  to  lose  its  elastic  properties,  and 
hence,  in  constructions  which  are  intended  to  last  for  any  length  of 
time,  the  members  should  be  so  designed  that  the  actual  stresses  lie 
well  below  the  elastic  limit. 

It  has  also  been  found  by  experiment  that,  for  iron  and  steel,  if 
the  stress  lies  well  within  the  elastic  limit,  it  can  be  removed  and 
repeated  indefinitely  without  causing  rupture ;  but  if  the  metal  is 
stressed  beyond  the  elastic  limit,  and  the  stress  is  repeated  or  alter- 
nates between  tension  and  compression,  it  will  eventually  cause 
rupture,  the  number  of  changes  necessary  to  produce  failure 
decreasing  as  the  difference  between  the  upper  and  lower  limits 
of  the  strain  increases.  This  is  known  as  the  fatigue  of  metals, 


6  RESISTANCE  OF  MATERIALS 

and  indicates  that  in  determining  the  resistance  of  any  material  the 
elastic  limit  is  much  more  important  than  the  ultimate  strength. 

Overstrain  of  any  kind  results  in  a  gradual  hardening  of  the 
material.  Where  this  has  already  occurred,  the  elastic  properties  of 
the  material  can  be  partially  or  wholly  restored  by  annealing ;  that 
is,  by  heating  the  metal  to  a  cherry  redness  and  allowing  it  to 
cool  slowly. 

6.  Working  stress.  The  stress  which  can  be  carried  by  any 
material  without  losing  its  elastic  properties  is  called  the  allowable 
stress  or  working  stress,  and  must  always  lie  below  the  elastic 
limit.  The  ratio  of  the  assumed  working  stress  to  the  ultimate 
strength  of  the  material  is  called  the  factor  of  safety ;  that  is, 

ultimate  strength 

(5)  Working  stress  = 

factor  of  safety 

Average  values  of  the  ultimate  strength,  factors  of  safety,  and  other 
elastic  constants  for  the  various  materials  used  in  construction  are 
given  in  Table  I. 

Since  for  wrought  iron  and  steel  the  elastic  limit  can  be  definitely 
located,  the  working  stresses  for  these  materials  is  usually  assumed 
as  a  certain  fraction,  say  i  to  |,  of  the  elastic  limit. 

Materials  like  cast  iron,  stone,  and  concrete  have  no  definite 
elastic  limits  ;  that  is,  they  do  not  conform  perfectly  to  Hooke's 
law.  For  such  materials,  therefore,  the  working  stress  is  usually 
assumed  as  a  small  fraction,  say  from  i  to  ^V'  °^  the  ultimate 
strength. 

Under  repeated  loads,  where  the  stress  varies  an  indefinite  num- 
ber of  times  between  zero  and  some  large  value,  the  working  stress 
may  be  assumed  as  f  of  its  value  for  a  static  load. 

If  the  stress  alternates  between  large  positive  and  negative 
values,  that  is,  between  tension  and  compression,  the  working  stress 
may  be  assumed  as  J  of  its  value  for  a  static  load. 

For  example,  if  the  elastic  limit  for  mild  steel  is  35,000  lb./in.2, 
the  working  stress  for  a  static  load  may  be  taken  as  18,000  lb./in.2; 
for  repeated  loads,  either  tensile  or  compressive,  as  12,000  lb./in.2; 
and  for  loads  alternating  between  tension  and  compression,  as 
6000  lb./in.2 


STRESS  AND  DEFORMATION 


ALLOWABLE  UNIT  STRESSES  IN  LB./IN.2 

(Also  called  working  stress  or  skin  stress) 

DEAD  LOAD 


ELASTIC 
LIMIT 

TENSION 

COMPRES- 
SION 

FLEXURE 

SHEAR 

TORSION 

Structural  steel 

32,000 

16,000 

16,000 

16,000 

12,000 

9,000 

Machinery  steel     . 

36,000 

18,000 

18,000 

18,000 

13,500 

13,500 

Crucible  O.H.  steel 

40,000 

20,000 

20,000 

20,000 

16,000 

16,000 

Cast  steel      .     . 

30,000 

12,000 

18,000 

15,000 

13,500 

10,000 

Wrought  iron    .     . 

25,000 

12,500 

12,500 

12,500 

9,000 

6,000 

Cast  iron 

4,500 

15,000 

7,500 

4,500 

4,500 

Phosphor  bronze    . 

20,000 

10,000 

10,000 

10,000 

8,000 

8,000 

Timber     .... 

3,000 

1,500 

1,000 

1,200 

120 

LIVE  LOAD 


Structural  steel 

10,600 

10,600 

10,600 

8,000 

6,000 

Machinery  steel     . 

12,000 

12,000 

12,000 

9,000 

9,000 

Crucible  O.H.  steel 

13,300 

13,300 

13,300 

10,600 

10,600 

Cast  steel      .     .     . 

8,000 

12,000 

10,000 

9,000 

6,600 

Wrought  iron    . 

8,400 

8,400 

8,400 

6,000 

4,000 

Cast  iron       .     .     . 

3,000 

10,000 

5,000 

3,000 

3,000 

Phosphor  bronze    . 

6,600 

6,600 

6,600 

5,400 

5.400 

Timber     .... 

1,000 

660 

800 

80 

REVERSIBLE  LOAD 


Structural  steel 

5300 

5300 

5300 

4000 

3000 

Machinery  steel  . 

6000 

6000 

6000 

4500 

4500 

Crucible  O.H.  steel 

6600 

6600 

6600 

5300 

5300 

Cast  steel   .  . 

4000 

6000 

5000 

4500 

3300 

Wrought  iron  . 

4200 

4200 

4200 

3000 

2000 

Cast  iron   .  .  . 

1500 

5000 

2500 

1500 

1500 

Phosphor  bronze  . 

3300 

3300 

3300 

2700 

2700 

Timber  .... 

500 

330 

400 

40 

8  RESISTANCE  OF  MATERIALS 

In  actual  practice  the  unit  working  stresses  are  usually  specified, 
and  the  designer  simply  follows  his  specifications  without  reference 
to  the  factor  of  safety.  Where  a  large  number  of  men  are  employed, 
this  method  eliminates  the  personal  equation  and  Insures  uniformity 
of  results.  The  student  should  become  familiar  with  both  methods, 
however,  as  it  is  no  small  part  of  an  engineer's  training  to  know 
what  relation  his  working  stress  should  bear  to  the  elastic  limit 
and  ultimate  strength  of  the  material. 

7.  Resilience.  The  work  done  in  straining  a  bar  up  to  the  elastic 
limit  of  the  material  is  called  the  resilience  of  the  bar.  The  area 
under  the  strain  diagram  from  the  origin  up  to  the  elastic  limit 
evidently  represents  the  work  done  on  a  unit  volume  of  the  material, 
say  one  cubic  inch,  in  straining  it  up  to  the  elastic  limit.  This  area 
therefore  represents  the  resilience  per  unit  volume,  and  is  called  the 
modulus  of  elastic  resilience  of  the  material. 

Thus,  if  p  denotes  the  unit  stress  and  s  the  unit  deformation  at 
the  elastic  limit,  then,  since  the  strain  diagram  up  to  this  point  is  a 
straight  line,  the  area  subtended  by  it,  or  modulus  of  elastic  resili- 

ence, is  i  ps.    Since  —  =  E,  the  expression  for  the  modulus  may 

o 

therefore  be  written 

2 
(6)  Modulus  of  elastic  resilience  = 


2  E 
Hence,  if  V  denotes  the  volume  of  the  test  piece,  its  total  resilience  is 

P2V 

(7)  Total  resilience  =  £  — 

Z  E 

The  resilience  of  a  bar  is  a  measure  of  its  ability  to  resist  a  blow 
or  shock  without  receiving  a  permanent  set.  If  a  load  W  is  applied 
gradually,  as  in  a  testing  machine,  the  maximum  stress  when  the 

W 

load  is  all  on  is  p  =  —  ,  where  A  denotes  the  area  of  the  cross 
A 

section  of  the  bar.  If,  however,  the  load  W  is  applied  suddenly, 
as  in  falling  from  a  height  ^,  it  produces  a  certain  deformation  of 
the  bar,  say  AZ,  and  consequently  the  total  external  work  done 
on  the  bar  is 

External  work  =  (h  +  AZ)  W. 


STRESS  AND  DEFORMATION  9 

If,  now,  the  unit  stress  p  produced  by  the  impact  lies  below  the 
elastic  limit,  the  total  internal  work  of  deformation  is 

•  Internal  work  =  |  (Ap)  AZ. 

In  the  case  of  a  suddenly  applied,  or  impact,  load,  like  that  due  to 
a  train  crossing  a  bridge  at  high  speed,  h  =  0,  and,  equating  the 
expressions  for  the  internal  and  external  work,  the  result  is 


2W 
whence  p  =  -- 

^\. 

Comparing  this  with  the  expression  for  the  stress  produced  by  a 
static  load,  namely,  p  —  —  ,  it  is  evident  that  a  suddenly  applied 

A 

load  produces  twice  the  stress  that  would  be  produced  by  the 
same  load  if  applied  gradually. 

8.  Poisson's  ratio.    Experiment  shows  that  when  a  bar  is  sub- 
jected to  tension  or  compression,  its  lateral,  or  transverse,  dimen- 
sions are  changed,  as  well  as  its  length.    Thus,  if  a  rod  is  pulled,  it 
increases  in  length  and  decreases  in  diameter  ;  if  it  is  compressed, 
it  decreases  in  length  and  increases  in  diameter. 

It  was  found  by  Poisson  that  the  ratio  of  the  unit  lateral  defor- 
mation to  the  unit   change  in  length  is  constant  for   any  given 

material.    This  constant  is  usually  denoted  by  —  ,  and  is  called  Poisj 

m 

son's  ratio.  Its  average  value  for  metals  such  as  steel  and  wrought 
iron  is  .3.  Thus,  suppose  that  the  load  on  a  steel  bar  produces  a 
certain  unit  deformation  s  lengthways  of  the  bar.  Then  its  unit 
lateral  deformation  will  be  approximately  .3s.  Hence,  the  total 
lateral  deformation  is  found  by  multiplying  this  unit  deformation 
by  the  width,  or  diameter,  of  the  bar. 

Values   of   Poisson's  ratio   for   various   materials   are  given   in 
Table  I. 

9.  Temperature   stress.    A  property  especially  characteristic   of 
metals  is  that  of  expansion  and  contraction  with  rise  and  fall  of 
temperature.    The  proportion  of  its  length  which  a  bar  free  to  move 
expands  when  its  temperature  is  raised  one  degree  is  called  its 


10  RESISTANCE  OF  MATERIALS 

coefficient  of  linear  expansion,  and  will  be  denoted  by  C.    Values 
of  this  constant  for  various  materials  are  given  in  Table  I. 

If  a  bar  is  prevented  from  expanding  or  contracting,  then  change 
in  temperature  produces  stress  in  the  bar,  called  temperature  stress. 
Thus,  let  I  denote  the  original  length  of  a  bar,  and  suppose  its 
temperature  is  raised  a  certain  amount,  say  T  degrees.  Then,  if  C 
denotes  the  coefficient  of  linear  expansion  for  the  material,  and  AZ 
the  amount  the  bar  would  naturally  lengthen  if  free  to  move, 

we  have 

AZ  =  CTl, 

and  consequently  the  unit  deformation  is  - 


Therefore,  if  p  denotes  the  unit  temperature  stress, 
(8)  p  =  sE 


APPLICATIONS 

1.  A  5-in.  copper  cube  supports  a  load  of  75  tons.    Find  its  change  in  volume. 
Solution.    Area  of  one  face  —  5  x  5  =  25  in.2    Unit  compressive  stress  p  on 

75  x  2000 
this  area  is  then  p  =  -  =  6000  lb./in.2   Modulus  of  elasticity  for  copper 

E  =  15,000,000  lb./in.2    Therefore  unit  vertical  contraction  s  =  —  =  -  ;  total 

E      2500 

vertical  contraction  AZ  =  J  •  s  =  5  •  --  =  —  in.   Since  Poisson's  ratio  for  copper 

1  340 

is  —  =  .340,  the  unit  deformation  laterally  is  .340  s  =  -  --  »  and  the  total  lateral 
m  2500 

.340 

deformation  is  5  x  -  -  =  .00068  in.    The  three  dimensions  of  the  deformed  cube 
2500 

are  therefore  5.00068,  5.00068,  4.998  ;  its  volume  is  124.984  in.3,  and  the  decrease 
in  volume  is  .016  in.3 

2.  A  f  -in.  wrought-iron  bolt  has  a  head  |  in.  deep.  If  a  load  of  4  tons  is  applied 
longitudinally,  find  the  factors  of  safety  in  tension  and  shear. 

Solution.   Area  of  body  of  bolt  at  root  of  thread  =  .442  in.2   Unit  tensile  stress 

in  bolt  is  p  =  4  X  200°  =  18,000  lb./in.2  Factor  of  safety  in  tension  =  5^2°.  =  2.7. 


Area  in  shear  =  TT  ----  =  1.47  in.2  Unit  shearing  stress  is  —  -  -  =  5440  lb./in.2 

40  000 

Factor  of  safety  in  shear  =  —  -  -  =  7.3. 
J  5440 


STRESS  AND  DEFOBMATION  11 

3.  A  steel  ring  fits  loosely  over  a  cylindrical  steel  pin  3  in.  in  diameter.    How 
much  clearance,  or  space  between  them,  should  there  be  in  order  that,  when  the 
pin  is  subjected  to  a  compressive  load  of  60  tons,  the  ring  shall  fit  tightly  ? 
Solution.    Unit  compressive  stress  in  pin  is 


=  17,000  Ib./in.* 
4 

Unit  longitudinal  deformation  s  =  —  =  -  -  -  --  =  .000566.    Unit  transverse  def- 

E      30,000,000 

ormation  =  .295  x  .000566  =  .000167.   Radial  clearance  =  1.5  x  .000167=  .00025  in. 

4.  A  post  1  ft.  in  diameter  supports  a  load  of  1  ton.  Assuming  that  the  stress  is 
uniformly  distributed  over  any  cross  section,  find  the  unit  normal  stress. 

5.  A  shearing  force  of  50  Ib.  is  uniformly  distributed  over  an  area  4  in.  square. 
Find  the  unit  shear. 

6.  A  steel  rod  500  ft.  long  and  1  in.  in  diameter  is  pulled  by  a  force  of  25  tons. 
How  much  does  it  stretch,  and  what  is  its  unit  elongation  ? 

7.  A  copper  wire  10  ft.  long  and  .04  in.  in  diameter  is  tested  and  found  to  stretch 
.289  in.  under  a  pull  of  50  Ib.    What  is  the  value  of  Young's  modulus  for  copper 
deduced  from  this  experiment  ? 

8.  A  round  cast-iron  pillar  18  ft.  high  and  10  in.  in  diameter  supports  a  load  of 
12  tons.    How  much  does  it  shorten,  and  what  is  its  unit  contraction  ? 

9.  A  wrought-iron  bar  20  ft.  long  and  1  in.  square  is  stretched  .266  in.    What  is 
the  force  acting  on  it  ? 

10.  What  is  the  lateral  contraction  of  the  bar  in  problem  9  ? 

11.  A  soft-steel  cylinder  1  ft.  high  and  2  in.  in  diameter  bears  a  weight  of 
40  tons.    How  much  is  its  diameter  increased  ? 

12.  A  copper  wire  100  ft.  long  and  .025  in.  in  diameter  stretches  2.16  in.  when 
pulled  by  a  force  of  15  Ib.    Find  the  unit  elongation. 

13.  If  the  wire  in  problem  12  was  250  ft.  long,  how  much  would  it  lengthen 
under  the  same  pull  ? 

14.  A  vertical  wooden  post  30  ft.  long  and  8  in.  square  shortens  .00374  in.  under 
a  load  of  half  a  ton.    What  is  its  unit  contraction  ? 

15.  How  great  a  pull  can  a  copper  wire  .2  in.  in  diameter  stand  without  breaking  ? 

16.  How  large  must  a  square  wrought-iron  bar  be  made  to  stand  a  pull  of 
3000  Ib.  ? 

17.  A  mild-steel  plate  is  ^  in.  thick.    How  wide  should  it  be  to  stand  a  pull  of 
10  tons  ? 

18.  A  round  wooden  post  is  6  in.  in  diameter.    How  great  a  load  will  it  bear  ? 

19.  A  wrought-iron  bar  is  20  ft.  long  at  32°  F.    How  long  will  it  be  at  95°  F.  ? 

20.  A  cast-iron  pipe  10  ft.  long  is  placed  between  two  heavy  walls.    What  will 
be  the  stress  in  the  pipe  if  the  temperature  rises  25°  ? 

21.  Steel  railroad  rails,  each  30  ft.  long,  are  laid  at  a  temperature  of  40°  F.  What 
space  must  be  left  between  them  in  order  that  their  ends  shall  just  meet  at  100°  F.  ? 

22.  In  the  preceding  problem,  if  the  rails  are  laid  with  their  ends  in  contact, 
what  will  be  the  temperature  stress  in  them  at  100°  F.  ? 

23.  A  f-in.  wrought-iron  bolt  failed  in  the  testing  machine  under  a  pull  of 
15,000  Ib.    Find  its  ultimate  tensile  strength. 


12 


RESISTANCE  OF  MATERIALS 


n 


n 


24.  Four  i-in.  steel  cables  are  used  with  a  block  and  tackle  on  the  hoist  of  a 
crane  whose  capacity  is  rated  at  6000  Ib.   What  is  the  factor  of  safety  ?   (Use  tables 
for  ultimate  strength  of  rope.) 

25.  A  vertical  hydraulic  press  weighing  100  tons  is  supported  by  four  2^-in. 
round  structural-steel  rods.     Find  the  factor  of  safety. 

26.  A  block  and  tackle  consists  of  six  strands  of  flexible  l-in.  steel  cable. 
What  load  can  be  supported  with  a  factor  of  safety  of  5  ? 

27.  A  vertical  wooden  bar  6  ft.  long  and  3  in.  in  diameter  is  found  to  lengthen 
.013  in.  under  a  load  of  2100  Ib.  hung  at  the  end.    Find  the  value  of  E  for  this  bar. 

28.  A  copper  wire  |  in.  in  diameter 
and  500  ft.  long  is  stretched  with  a  force 
of  100  Ib.  when  the  temperature  is  80°  F. 
Find  the  pull  in  the  wire  when  the  tem- 
perature is  0°F.,  and  the  factor  of  safety. 

29.  An  extended  shank  is  made  for 
a  -i^-in.  drill  by  boring  a  ^|-in.  hole  in 

FIG.  3  the  end  of  a  piece  of  |-in.  cold-rolled 

steel,  fitting  the  shank  into  this,  and 

putting  a  steel  taper  pin  through  both  (Fig.  3).  Standard  pins  taper  1  in.  per  foot. 
What  size  pin  should  be  used  in  order  that  the  strength  of  the  pin  against  shear 
may  equal  the  strength  of  the  drill  shank  in  compression  around  the  hole  ? 

30.  The  head  of  a  steam  cylinder  of  12-in.  inside  diameter  is  held  on  by  ten 
wrought-iron  bolts.    How  tight  should  these  bolts  be  screwed  up  in  order  that 
the  cylinder  may  be  steam  tight  under  a  pres- 
sure of  180  lb./in.2  ? 

31.  Find  the  depth  of  head  of  'a  wrought- 
iron  bolt  in  terms  of  its  diameter  in  order  that 
the  tensile  strength  of  the  bolt  may  equal  the 
shearing  strength  of  the  head. 

32.  The  pendulum  rod  of  a  regulator  used  in 
an  astronomical  observatory  is  made  of  nickel 
steel  in  the  proportion  of  35.7  per  cent  nickel  to 
64.3  per  cent  steel.    The  coefficient  of  expansion 
of  this  alloy  is  approximately  0.0000005. 

The  rod  carries  two  compensation  tubes,  A 
and  B  (Fig.  4),  one  of  copper  and  the  other  of 
alloy,  the  length  of  the  two  together  being  10  cm. 
If  the  length  of  the  rod  to  the  top  of  tube  A  is 
1m.,  find  the  lengths  of  the  two  compensation 
tubes  so  that  a  change  in  temperature  shall  not 
affect  the  length  of  the  pendulum. 

33.  Refer  to  the  Watertown  Arsenal  Reports 
(United  States  Government  Reports  on   Tests  of 

Metals).,  and  from  the  experimental  results  there  tabulated  draw  typical  strain  dia- 
grams for  mild  steel,  wrought  iron,  cast  iron,  and  timber,  and  compute  E  in  each  case. 

34.  A  steel  wire  \  in.  in  diameter  and  a  brass  wire  1  in.  in  diameter  jointly 
support  a  load  of  1200  Ib.    If  the  wires  were  of  the  same  length  when  the  load 
was  applied,  find  the  proportion  of  the  load  carried  by  each. 


FIG.  4 


STEESS  AND  DEFORMATION  13 

35.  An  engine  cylinder  is  10  in.  inside  diameter  and  carries  a  steam  pressure 
of  801b./in.-    Find  the  number  and  size  of  the  bolts  required  for  the  cylinder 
head  for  a  working  stress  in  the  bolts  of  2000  lb./in.2 

36.  Find  the  required  diameter  for  a  short  piston  rod  of  hard  steel  for  a  piston 
20  in.  in  diameter  and  steam  pressure  of  125  lb./in.2    Use  factor  of  safety  of  8. 

37.  A  rivet  i  in.  in  diameter  connects  two  wrought-iron  plates  each  '£  in.  thick. 
Compare  the  shearing  strength  of  the  rivet  with  the  crushing  strength  of  the  plates 
around  the  rivet  hole. 

38.  In  the  United  States  government  tests  of  rifle-barrel  steel  it  was  found  that 
for  a  certain  sample  the  unit  tensile  stress  at  the  elastic  limit  was  71, 000  lb./in.2, 
and  that  the  ultimate  tensile  strength  was  118,000  lb./in.2   What  must  the  factor  of 
safety  be  in  order  to  bring  the  working  stress  within  the  elastic  limit  ? 

39.  In  the  United  States  government  tests  of  concrete  cubes  made  of  Atlas 
cement  in  the  proportions  of  1  part  of  cement  to  3  of  sand  and  6  of  broken  stone, 
the  ultimate  compressive  strength  of  one  specimen  was  883  lb./in.2,  and  of  another 
specimen  was  3256  lb./in.2   If  the  working  stress  is  determined  from  the  ultimate 
strength  of  the  first  specimen  by  using  a  factor  of  safety  of  5,  what  factor  of  safety 
must  be  used  to  determine  the  same  working  stress  from  the  other  specimen  ? 

40.  An  elevator  cab  weighs  3  tons.    With  a  factor  of  safety  of  5,  how  large 
must  a  steel  cable  be  to  support  the  cab  ? 

41.  A  hard-steel  punch  is  used  to  punch  holes  in  a  wrought-iron  plate  |  in. 
thick.    Find  the  diameter  of  the  smallest  hole  that  can  be  punched. 

42.  A  mild-steel  plate  10  in.  square  and  ^in.  thick  is  stretched  0.002  in.  in  one 
direction  by  a  certain  pull.   What  pull  must  be  applied  at  right  angles  to  reduce 
the  first  stretch  to  0.0014  in.  ? 

43.  A  structural  steel  tie  rod  of  a  bridge  is  to  be  25  ft.  long  when  the  bridge  is 
completed.    What  should  its  original  length  be  if  the  maximum  stress  in  it  when 
loaded  is  18,000  lb./in.2  ? 

44.  A  hard-steel  punch  is  used  to  punch  a  circular  hole  ^  in.  in  diameter  in  a 
wrought-iron  plate  ^  in.  thick.    Find  the  factor  of  safety  for  the  punch  when  in  use. 

45.  A  cast-iron  flanged  shaft  coupling  is  bolted  together  with  1-in.  wrought- 
iron  bolts,  the  distance  from  the  axis  of  each  bolt  to  the  axis  of  the  shaft  being  6  in. 
If  the  shaft  transmits  a  maximum  torque  of  12,000  ft.-lb.,  find  the  number  of 
bolts  required. 

46.  A   steam  cylinder  of  16  in.  inside  diameter  carries  a  steam  pressure  of 
150  lb./in.2   Find  the  proper  size  for  the  hard-steel  piston  rod,  and  the  number  of 
5-in.  wrought-iron  bolts  required  to  hold  on  the  cylinder  head. 

47.  A  horizontal  beam  10ft.  long  is  suspended  at  one  end  by  a  wrought-iron  rod 
12  ft.  long  and  i  in.  in  diameter,  and  at  the  other  end  by  a  copper  rod  12  ft.  long 
and  1  in.  in  diameter.    At  what  point  on  the  beam  should  a  load  be  placed  if  the 
beam  is  to  remain  horizontal ;  that  is,  if  each  rod  is  to  stretch  the  same  amount  ? 

48.  When  a  bolt  is  screwed  up  by  means  of  a  wrench,  the  tension  Tin  the  bolt  in 
terms  of  the  pull  P  on  the  handle  of  the  wrench  is  found  to  be  given  approximately 
by  the  empirical  formula  T  —  75  P 

for  a  wrench  of  maximum  length  of  from  15  to  16  times  the  diameter  of  the  bolt. 
What  is  the  largest  wrench  that  should  be  used  on  a  £-in.  wrought-iron  bolt,  and 
what  is  the  maximum  pull  that  should  be  exerted  on  the  handle  ? 


14 


RESISTANCE  OF  MATERIALS 


49.  For  a  steam-tight  joint  the  pitch  (or  distance  apart)  of  studs  or  bolts  in 
cylinder  heads  is  determined  by  the  empirical  formulas 

High-pressure  cylinders,  pitch  =  3.5d, 

Intermediate-pressure  cylinders,          pitch  =  4.5  d, 
Low-pressure  cylinders,  pitch  =  5.5  d  ; 

or,  in  general,  pitch  = 

where  d  =  diameter  of  studs  or  bolts, 

t  =  thickness  of  head  or  cover  in  sixteenths  of  an  inch, 
w  =  steam  pressure  in  lb./in.2 

Calculate  the  number,  size,  and  pitch  of  steel  studs  for  a  steam  cylinder  20  in. 
inside  diameter  under  a  pressure  of  150  lb./in.2  (high  pressure). 


FIG.  5 

50.  The  standard  proportions  for  a  cott-ered  joint  with  wrought-iron  rods  and 
steel  cotter  of  the  type  shown  in  Fig.  5  are  as  indicated  on  the  figure.  Show  that 
these  relative  proportions  make  the  joint  practically  of  uniform  strength  in  tension, 
compression,  and  shear. 


SECTION  II 


FIRST  AND  SECOND  MOMENTS 

10.  Static  moment.  If  a  force  acts  upon  a  body  having  a  fixed 
axis  of  rotation,  it  will  in  general  tend  to  produce  rotation  of  the 
body  about  this  axis.  This  tendency  to  rotate  becomes  greater  as 
the  magnitude  of  the  force  increases,  and  also  as  its  distance  from 
the  axis  of  rotation  increases.  The  numerical  amount  of  this  tend- 
ency to  rotate  is  thus  measured  by  the  product  of  the  force  by  its 
perpendicular  distance  from  the  given  axis,  or  center  of  rotation. 
This  product  is  called  the  first  moment,  or  static  moment,  of  the 
force  with  respect  to  the  given  axis,  or  point. 

Thus,  let  F  denote  any  force,  P  the  fixed  axis  of  rotation,  as- 
sumed to  be  at  right  angles  to  the  plane  of  the  paper,  and  d  the 
perpendicular  distance  of  F  from  P.  Then  d  is  called  the  lever  arm 
of  the  force,  and  its  moment  about  P  is  defined  as 

Moment  =  force  X  lever  arm, 

or,  if  the  moment  is  denoted  by  M, 


(9) 


M  =  Fd. 


P 

(Center  or  axis 
of  rotation) 


It  is  customary  to  call  the 
moment  positive  if  it  tends  to 
produce  rotation  in  a  clockwise 
direction,  and  negative  if  its 
direction  is  counter-clockwise 
(Fig.  6). 

11.  Fundamental  theorem  of 
moments.  When  two  concur- 
rent forces  act  on  a  body  simul- 
taneously, their  joint  effect  is 
the  same  as  that  of  a  single  force,  given  in  magnitude  and  direction 
by  the  diagonal  of  the  parallelogram  formed  on  the  two  given 

15 


FIG.  6 


16 


RESISTANCE  OF  MATERIALS 


N 


W 


0  W 

FIG.  7 


forces  as  adjacent  sides  (Fig.  7).  This  single  force,  which  is  equiv- 
alent to  the  two  given  forces,  is  called  their  resultant. 

Any    number    of    concurrent 
forces  may  be  thus  combined  by 
\  finding  the  resultant  of  any  two, 

\XP  combining  this  with  the   third, 

\  etc.    Or,  what  amounts  to  the 

\  same  thing,  the  given  forces  may 

\         be  placed  end  to  end,  forming  a 

1      polygon,  and  their  final  resultant 

will  then  be  the  closing  side  of 
this  polygon  (Fig.  8). 
Now,  in  Fig.  9,  let  F1  and  F2  be  any  two  concurrent  forces,  and  F 
their  resultant.  Also  let  0  be  any  given  point,  and  0^  00,  c£,  the  angles 
between    OA  and  the  forces 
F^,  Ff  F,  respectively.  Then, 
taking  moments  about  0, 

Moment  of  Fl  about  0 

=  Flx  OAsin  Qv 
Moment  of  Fz  about  0 

=  F0  x  OA  sin  02. 

The  sum  of  these  moments  is 

=  FlxOA  sin  0l  +  F2  x  OA  sin  #2  =  OA  (^  sin  0l  +  F2  sin  02). 

But,  since  F  is  the   resultant 
of  Fl  and  F# 

F  sin  <£  =Fl  sin  0l+F2  sin  0^ 
and  consequently 

2}  M=OA  xFsm(j>. 

The  right  member,  however,  is 
the  moment  of  the   resultant 

\^~---A  F  with  respect  to  0.  Therefore, 

FlG<9  since  0  is  arbitrary,   the  sum 

of  the  moments  of  any  two  concurrent  forces  with  respect  to  a 
given  point  is  equal  to  the  moment  of  their  resultant  with  respect 
to  this  point. 


FIRST  AND  SECOND  MOMENTS 


17 


If  the  forces  Fl  and  F2  are  parallel,  introduce  two  equal  and 
opposite  forces  H,  —  H,  as  shown  in  Fig.  10,  and  combine  the  iTs 
with  Fl  and  F2  into  resultants  F^,  F'r  Transferring  these  resultants 
FV  FZ  to  their  point  of  intersection  0,  they  may  now  be  resolved 
into  their  original  components,  giving  two  equal  and  opposite  forces, 
-f-  H  and  —  H,  which  cancel,  and  a  resultant  Fl  +  F2  parallel  to  Fl 
and  F2. 

Moreover,  applying  the  theorem  of  moments  proved  above  to  the 
concurrent  forces  F^,  F%  at  0,  the  sum  of  their  moments  about  any 
point  is  equal  to  the  moment 
of  their  resultant  Fl  +  F2 
about  the  same  point.  But 
the  moment  of  F[  is  equal 
to  the  sum  of  the  moments 
of  Fl  and  —  H,  and,  simi- 
larly, the  moment  of  F^  is 
equal  to  the  sum  of  the 
moments  of  Fz  and  +  H. 
Since  the  forces  +  H  and 
-  H  have  the  same  line 
of  action,  their  moments 
about  any  point  cancel,  and  therefore  the  theorem  of  moments  is 
also  valid  for  parallel  forces. 

This  theorem  may  obviously  be  extended  to  any  number  of  forces 
by  combining  the  moments  of  any  two  of  them  into  a  resultant 
moment,  combining  this  resultant  moment  with  the  moment  of  the 
third  force,  etc.  Hence, 

The  sum  of  the  moments  of  any  number  of  forces  lying  in  the  same 
plane  with  respect  to  a  given  point  in  this  plane  is  equal  to  the  moment 
of  their  resultant  with  respect  to  this  point. 

12.  Center  of  gravity.  An  important  application  of  the  theorem 
of  moments  arises  in  considering  a  system  of  particles  lying  in  the 
same  plane  and  rigidly  connected.  The  weights  w^  w^  •  •  •,  wn  of  the 
particles  are  forces  directed  toward  the  center  of  the  earth.  Since 
this  is  relatively  at  an  infinite  distance  as  compared  with  the  dis- 
tances between  the  particles,  their  weights  may  be  regarded  as  a 
system  of  parallel  forces. 


FIG. 10 


..V 


18  RESISTANCE  OF  MATERIALS 

The  total  weight  W  of  all  the  particles  is 

W=w  +  ivz+  .  .  .  4-  wn  =       w  ; 


that  is,  W  is  the  resultant  of  the  n  parallel  forces  w^  w^  •  •  .,  wn. 
The  location  of  this  resultant  W  may  be  determined  by  applying 
the  theorem  of  moments.  Thus,  let  x^  #2,  •  •  -,  xn  denote  the  perpen- 
dicular distances  of  w^  w0,  •  •  •,  wn  from 
any  fixed  point  0  (Fig.  11).  Then,  if 
XQ  denotes  the  perpendicular  distance 
of  the  resultant  W  from  0,  by  the 
theorem  of  moments 


whence 
FIG.  11 


2'o-      w     > 


or,  since  W  =  ^  w,  this  may  also  be  written 

(10)  «.=4^- 

7,  iv 

This  relation  determines  the  line  of  action  of  W  for  the  given 
position  of  the  system.  If,  now,  the  system  is  turned  through  any 
angle  in  its  plane,  and  the  process  repeated,  a  new  line  of  action 
for  W  will  be  determined.  The  point  of  intersection  of  two  such 
lines  is  called  the  center  of  gravity  of  the  system.  From  the  method 
of  determining  this  point  it  is  evident  that  if  the  entire  weight  of 
the  system  was  concentrated  at  its  center  of  gravity,  this  single 
weight,  or  force,  would  be  equivalent  to  the  given  system  of  forces, 
no  matter  what  the  position  of  the  system  might  be. 

If  the  particles  do  not  all  lie  in  the  same  plane,  a  reference  plane 
must  be  drawn  through  0  instead  of  a  reference  line.    In  this  case 

">-\ 

the  equation  XQ  =  ^       determines  the  position  of  a  plane  in  which 

the  resultant  force  W  must  lie.  The  intersection  of  three  such 
planes  corresponding  to  different  positions  of  the  system  of  particles 
will  then  determine  a  point  which  is  the  required  center  of  gravity. 


FIRST  AND  SECOND  MOMENTS  19 

If  m^  m^  -  •  .,  mn  denote  the  masses  of  the  n  particles,  and  M  their 
sum,  then,  since 

W=Mg,     Wl  =  m^     w2  =  m2g,     •  •  •,    wn  =  mng, 

where  g  denotes  the  acceleration  due  to  gravity,  the  above  relations 
for  determining  the  center  of  gravity  become 


or,  since  g  is  constant, 
(11)  M 


ill 


The  point  determined  from  these  relations  by  taking  the  system  of 
particles  in  two  or  more  positions  is  called  the  center  of  inertia  or 
center  of  mass.  Since  these  relations  are  identical  with  those  given 
above,  it  is  evident  that  the  center  of  mass  is  identical  with  the 
center  of  gravity. 

13.  Centroid.  It  is  often  necessary  to  determine  the  point  called 
the  center  of  gravity  or  center  of  mass  without  reference  to  either 
the  mass  or  weight  of  the  body,  but  simply  with  respect  to  its 
geometric  form. 

For  a  solid  body  let  Av  denote  an  element  of  volume,  Am  its 
mass,  and  D  the  density  of  the  body.  Then,  since  mass  is  jointly 
proportional  to  volume  and  density, 


Therefore  the  formulas  given  above  may  be  written 


or,  since  the  density  D  is  constant,  these  become 

(12)      '  r 


Since  the  point  previously  called  the  center  of  gravity  or  center  of 
mass  is  now  determined  simply  from  the  geometric  form  of  the 
body,  it  is  designated  by  the  special  name  centroid. 


20 


RESISTANCE  OF  MATERIALS 


Evidently  it  is  also  possible  to  determine  the  centroid  of  an  area 
or  line,  although  neither  has  a  center  of  gravity  or  center  of  mass, 
since  mass  and  weight  are  properties  of  solids. 

For  a  plane  area  the  centroid  is  determined  by  the  equations 


(13) 


where  A#  denotes  an  element  of  area  and^l  the  total  area  of  the  figure. 
Similarly,  for  a  line  or  arc  the  centroid  is  given  by 


(14) 


where  A?  denotes  an  element  of  length  and  L  the  total  length  of 

the  line  or  arc. 

14.  Centroid  of  triangular  area.   To  find  the  centroid  of  a  triangle, 

divide  it  up  into  narrow  strips  parallel  to  one  side  AC  (Fig.  12). 

Since  the  centroid  of  each  strip  PQ  is  at 
its  middle  point,  the  centroid  of  the  en- 
tire figure  must  lie  somewhere  on  the 
line  BD  joining  these  middle  points  ;  that 
is,  on  the  median  of  the  triangle.  Simi- 
larly, by  dividing  the  triangle  up  into 
strips  parallel  to  another  side  BC,  it  is 
proved  that  the  centroid  must  also  lie 
on  the  median  AE.  The  point  of  inter- 
section G  of  these  two  medians  must 
therefore  be  the  centroid  of  the  triangle. 

Since  the  triangles  DEG  and  ABG  are  similar, 

DG      DE 


and  since  DE  =  1  A  B,  this  gives 


The  centroid  of  a  triangle  therefore  lies  on  a  median  to  any  side  at 
a  distance  of  one  third  the  length  of  the  median  from  the  opposite 
vertex.  From  this  it  also  follows  that  the  perpendicular  distance 


FIRST  AND  SECOND  MOMENTS 


21 


of  the  centroid  G  from  any  side  is  one  third  the  distance  of  the 
opposite  vertex  from  that  side. 

15.  Centroid  of  circular  arc.  For  a  circular  arc  CD  (Fig.  13)  the 
centroid  G  must  lie  on  the  diameter  OF  bisecting  the  arc.  Now 
suppose  the  arc  divided  into  small  segments,  and  from  the  ends 
of  any  segment  PQ  draw  PR 
parallel  to  the  chord  CZ>,  and 
QR  perpendicular  to  this  chord. 
Since  the  moment  of  the  entire 
arc  with  respect  to  a  line  AB 
drawn  through  0  perpendicular 
to  OF  must  be  equal  to  the  sum 
of  the  moments  of  the  small  seg- 
ments PQ  with  respect  to  this 
line,  the  equation  determining 
the  centroid  is 


X  x. 


FIG. 13 


But  from  the  similarity  of  the  triangles  PQR  and  OQE  we  have 

PQ      OQ      r 


PQ  -  x  =       PR 


Therefore  PQ  -  x  =  PR  -  r,  and  consequently 
or,  since  the  radius  r  is  constant, 


^PQ.  x  =  r^  PR  =  r  .  chord  CD. 
The  position  of  the  centroid  is  therefore  given  by 


(15) 


chord 
arc 


.  radius. 


If  the  central  angle  COD  is  denoted  by  2  a,  then  arc=2ra  and 
chord  =  2  r  sin  a,  and  therefore  the  expression  for  the  centroid  may 
be  written 


For  a  semicircle  2  a  =  TT,  and  consequently 


22 


RESISTANCE  OF  MATERIALS 


16.  Centroid  of  circular  sector  and  segment.  To  determine  the  cen- 
troid  of  a  circular  sector  OCB  (Fig.  14),  denote  the  radius  by  r  and 

the  central  angle  COB  by  2  a. 
Then  any  small  element  OPQ 
of  the  sector  may  be  regarded 
as  a  triangle  the  centroid  of 
which  is  on  its  median  at  a 
distance  of  Jr  from  0.  The 
centroids  of  all  these  elemen- 
tary triangles  therefore  lie  on 
a  concentric  arc  DEF  oi  radius 
|  r,  and  the  centroid  of  the 
entire  sector  coincides  with 
the  centroid  of  this  arc  DEF.  Therefore,  from  the  results  of  the  pre- 
ceding article,  the  centroid  of  the  entire  sector  OCB  is  given  by 


FIG. 14 


(16) 


2  sin  a 

=  —  r 

3  a 


For  a  semicircular  area  of  radius  r  the  distance  of  the  centroid 
from  the  diameter,  or  straight  side,  is 


37T 

To  determine  the  centroid  of  a  circular  segment  CBD  (Fig.  15), 
let  G  denote  the  centroid  of 
the  entire  sector  OCBD,  GQ 
of  the  segment  CBD,  and  G1 
of  the  triangle  OCD.  Then  the 
position  of  GQ  may  be  deter- 
mined by  noting  that  the  sum 
of  the  moments  of  the  triangle 
OCD  and  the  segment  CBD 
about  any  point,  say  0,  is  equal 
to  the  moment  of  the  entire 
sector  about  this  point.  Thus,  FlG' 15 

if  AQ,  A^  A  denote  the  areas  of  the  segment,  triangle,  and  sector,  re- 
spectively, and  XQ,  x^  x,  the  distances  of  their  centroids  from  0,  then 


whence 


FIRST  AND  SECOND  MOMENTS 

Ax  -  A 


Now  let  c  denote  the  length  of  the  chord  CD  and  a  the  length  of  the 
arc  CBD.   Then,  from  the  results  of  this  and  the  preceding  articles, 


2  re 
-— 

3  a 


2       *     <? 
*i  =  8Y-4' 


and  also,  from  geometry, 

1  1 

A  =  -ar,          Ai  =  2 

Inserting  these  values  in  the  expression  for  #0,  the  result  is 

Trr*  °       ^A° 
For  a  semicircle,  AQ  =  —  and  c  =  2  r.    Therefore,  in  this  case,  as 

also  shown  above, 


17.  Centroid  of  para- 

u 

bolic    segment.     For    a 
parabolic  segment  with       _i^ 
vertex   at  A   (Fig.  16) 
the  position  of  the  cen- 
troid G  is  given  by 

(18)  *.  =  !« 


where  a  and  b  denote  the  sides  of  the  circumscribing  rectangle.  Also, 


FIG.  16 


(19) 


Area^LBC  =  -  ab. 
3 


For  the  external  segment  ABD  (Fig.  16)  the  centroid  is  given  by 

3  3, 

(20)  CCQ  =  —  a,          y0  =  T  &> 

and  the  area  of  the  external  segment  is 


(21) 


Area  ABD  =  -  ab. 
3 


24 


RESISTANCE  OF  MATERIALS 


FIG. 17 


18.  Axis  of  symmetry.   If  a  figure  has  an  axis  of  symmetry,  then 
to  any  element  of  the  figure  on  one  side  of  the  axis  there  must 

correspond*  an  equidistant 
element  on  the  opposite  side, 
and  since  the  moments  of 
these  equal  elements  about 
the  axis  of  symmetry  are 
equal  in  amount  and  oppo- 
site in  sign,  their  sum  is 
zero  (Fig.  17).  Since  the 
moment  of  each  pair  of  ele- 
ments with  respect  to  the  axis 
of  symmetry  is  identically 
zero,  the  total  moment  is  also  zero,  and  hence  the  centroid  of  the 
figure  must  lie  on  the  axis  of  symmetry. 

When  a  figure  has  two  or  more  axes  of  symmetry,  their  inter- 
section completely  determines  the  centroid. 

19.  Centroid  of  composite  figures.     To    determine  the   centroid 
of  a  figure  made  up  of  several  parts,  the  centroid  of  each  part 
may    first  be    determined    separately.     Then,    assuming    that    the 
area  of  each  part  is  concentrated  at  its  centroid,  the  centroid  of 
the    entire    figure    may    be    deter- 
mined by  equating  its  moment  to 

the    sum    of   the   moments    of   the 
several  parts. 

To  illustrate  this  method,  let  it 
be  required  to  find  the  centroid  of 
the  I -shape  shown  in  Fig.  18.  Since 
the  figure  has  an  axis  of  symmetry 
MN,  the  centroid  must  lie  some- 
where on  this  line.  To  find  its 
position,  divide  the  /  into  three 
rectangles,  as  indicated  by  the 
dotted  lines  in  the  figure.  The 


\M 


I 

CKJ) 
..}__ 

i 
i 


\N 
FIG. 18 


centroids  of  these  rectangles  are  at  their  centers  a,  5,  c.  Therefore, 
denoting  these  three  areas  of  the  rectangles  by  A,  B,  C,  respectively, 
and  taking  moments  with  respect  to  the  base  line,  the  distance  of  the 


FIRST  AND  SECOND  MOMENTS  25 

centroid  of  the  entire  figure  from  the  base  is  found  to  be 
_A  x  ad+  B  x  bd+  C  X  cd 

/£•     ___  . ^ 

As  another  example,  consider  the  circular  disk  with  a  circular 
hole  cut  in  it,  shown  in  Fig.  19.  Here  also  the  centroid  must  lie 
somewhere  on  the  axis  of  symmetry  (71?  C2.  Therefore,  denoting 

the  radii  of  the  circles  by 
R,  r,  as  shown,  and  taking 
moments  about  the  tangent 
perpendicular  to  the  line  of 
centers,  the  distance  XQ  of 
the  centroid  from  this  tan- 
gent is  found  to  be 


or,  since  #2  =  R  and  xl  =  R  —  e,  where 
the  hole,  or  distance  between  centers, 


TrR'2  —  ?rr2 
denotes  the  eccentricity  of 


R3  -r\R-  e) 


20.  Moment  of  inertia.    In  the  analysis  of  beams,  shafts,  and 
columns  it  will  be  found  necessary  to  compute  a  factor,  called  the 
moment  of  inertia,  which  depends  only  on  the  shape  and  size  of  the 
cross   section    of    the   member. 
This    shape    factor    is    usually 
denoted  by  7,  and  is  defined  as 
the  sum  of  the  products  obtained 

by  multiplying  each  element  of     

area  of  the  cross  section  by  the 

square   of  its   distance  from  a 

given   line   or  point.    Thus,  in 

Fig.  20,  if  A^4  denotes  an  element  of  area  and  y  its  distance  from  any 

given  axis  00,  then  the  moment  of  inertia  of  the  figure  with  respect 

to  this  axis  is  defined  as 


FIG. 20 


26 


RESISTANCE  OF  MATERIALS 


Since  an  area  is  not  a  solid  and  therefore  does  not  possess  inertia, 
the  shape  factor  /  should  not  be  called  moment  of  inertia,  but  rather 
the  second  moment  of  area,  since  the  distance  y  occurs  squared. 

To  compute  I  for 
any  plane  area,  divide 
the  area  up  into  small 
elements  A^  (Fig.  21). 
-H  ]  Then  the  first  (or 
static)  moment  of  each 
element  with  respect 
to  any  axis  00  is  yA^4, 
where  y  denotes  the 
distance  of  this  ele- 
ment from  the  given 
axis.  Now  erect  on  A^4 
as  base  a  prism  of 

height  y.  If  this  is  done  for  every  element  of  the  plane  area,  the 
result  will  be  a  solid,  or  truncated  cylinder,  as  shown  in  Fig.  21, 
the  planes  of  the  upper  and  lower  bases  intersecting  in  the  axis 
00  at  an  angle  of  45°. 

Let  V  denote  the  volume  of  this  moment  solid,  as  it  will  be  called, 
and  y  the  distance  of  its  centroidal  axis  from  00.  Then,  by  the 
theorem  of  moments, 


\ 


FIG.  21 


Since  A  V  =  #A A,  the 
right  member  becomes 


Hence 

(22)     I  =  Ft/0. 

21.    I    for    rectangle.  FIG.  22 

Let   it   be   required   to 

find  I  for  a  rectangle  of  breadth  b  and  height  h  with  respect  to  an 
axis  through  its  centroid,  or  middle  point,  and  parallel  to  the  base 
(Fig.  22).  The  moment  solid  in  this  case  consists  of  a  double  wedge, 


FIRST  AND  SECOND  MOMENTS 


27 


as  shown  in  Fig.  22,  the  base  of  each  wedge  being  — ,  its  height  - , 
and  its  volume 

V  =  -  base  x  altitude  =  —  • 
2  8 

Since  the  centroid  of  a  triangular  wedge,  like  that  of  a  triangle,  is 
at  a  distance  of  |  its  altitude  from  the  vertex, 

_2      h_h 
^°~3  X  2~3* 

Therefore  1=2  Vii^  =  -—  • 


For  any  plane  area  the  /'s  with  respect  to  two  parallel  axes  are 
related  as  follows : 

Let    00   denote    an    axis 
through  the  centroid  of  the 

figure,  AA  any  parallel  axis,     _O 

and  d  their  distance  apart 
(Fig.  23).  Also  let  I0  denote 
the  /  of  the  figure  with  re- 
spect to  the  axis  00,  and  IA 
with  respect  to  the  axis  AA. 
Then,  from  the  definition  of  7, 


FIG. 23 


But  since  00  is  a  centroidal  axis,  ^ykA  =  0  for  this  axis.    There- 
fore, since  V«/2A^4  =  /0,  the  above  expression  becomes 

(23)  I A  =  TO  +  <?A. 

From  this  relation  it  is  evident  that  the  /  for  a  centroidal  axis  is 
less  than  for  any  parallel  axis. 

As  an  application  of  this  formula,  find  the  I  for  a  rectangle  with 

-j   -i  q  7 

respect  to  its  base.    From  what  precedes,  I0=  —  •  Also,  d=  -  and 

I  L-  2 

A  =  bh.    Hence  the  /  for  a  rectangle  with  respect  to  its  base  is 


3 


28 


RESISTANCE  OF  MATEEIALS 


22.  I  for  triangle.  Consider  a  triangle  of  base  b  and  altitude  h 
and  compute  first  its  /with  respect  to  an  axis  A  A  through  its  vertex 
and  parallel  to  the  base  (Fig.  24).  The  moment  solid  in  this  case 
is  a  pyramid  of  base  bh  and  altitude  A,  the  volume  of  which  is 

V=-  base  x  altitude  =  —  • 
o  o 

Since  the  centroid  of  this  pyramid  is  at  a  distance  ?/0  =  ^  h  from 
the  vertex,  we  have  3 


FIG. 24 


To  find  /for  the  triangle  with  respect  to  an  axis  00  through  its 
centroid  and  parallel  to  AA,  apply  the  theorem 


bh*  bh         ,  ,      2 

Since  in  the  present  case  IA  =  —  ,  ^4  =  —  -,  and  d  =  -  h,  we  have 

therefore 


Similarly,  for  the  axis  ^^  we  have 


r 
1B  — 


23.  I  for  circle.  In  computing  the  /for  a  circle,  it  is  convenient 
to  determine  it  first  with  respect  to  an  axis  through  the  center  of 
the  circle  and  perpendicular  to  its  plane  (the  so-called  polar 
moment  of  inertia  of  the  circle)  . 


FIRST  AND  SECOND  MOMENTS 


29 


Consider  the  circle  as  made  up  of  a  large  number  of  elementary 
triangles  OAB  with  common  vertex  at  0  (Fig.  25).  Since  the  alti- 
tude of  each  of  these  triangles  is  the  radius  R  of  the  circle,  from 
the  preceding  article  the  /  for  each  with  respect  to  the  point  0  is 

—  --    For  the  entire  circle,  therefore, 


or,  since    T  AB  =  circumference  =  2  irR,  this  becomes 


(24) 


If  D  denotes  the  diameter  of  the  circle,  then  R  =  -~  and  we  also  have 


(35)  *>-• 

If  XX  and  YY  are  two  rectan- 
gular diameters  of  the  circle,  and  r 
is  the  distance  of  any  element  of 
area  AA  from  their  point  of  inter- 
section 0  (Fig.  25),  then 


I0  =  IY 


Hence 

(26) 

Since  a  circle  is  symmetrical  about  all  diameters,  we  have  Ix  =  IY. 
Therefore  the  I  of  a  circle  with  respect  to  any  diameter  is 


or 


(27) 


7TD* 

64 


24.  I  for  composite  figures.  When  a  plane  figure  can  be  divided 
into  several  simple  figures,  such  as  triangles,  rectangles,  and  circles, 
the  /  of  the  entire  figure  with  respect  to  any  axis  may  be  found  by 
adding  together  the  J's  for  the  several  parts  with  respect  to  this 
axis.  Thus,  in  Fig.  26  each  area  may  be  regarded  as  the  difference 


30 


RESISTANCE  OF  MATERIALS 


of  two  rectangles  —  a  large  rectangle  of  base  B  and  height  77,  and  a 
smaller  rectangle  of  base  b  and  height  h.    Consequently  the  /  for 

either  figure,  with 
respect  to  its  cen- 
troidal  axis  GG, 


in, 


is  given  by 


W 
12' 


12 

Similarly,  the 
figures  shown  in 
Fig.  27  may  each 
be  regarded  as  the 
sum  of  two  rect- 
angles, and  hence 
the  /  for  either 
of  these  figures  with  respect  to  its  centroidal  axis  GG  is  given  by 


FIG.  26 


FIG. 27 


1  = 


BH 


"h 


12        !2 


For  the  angle,  or  tee,  shown  in  Fig.  28  the  7"  about  the  base  line 
00  is  the  sum  of  the  J's  for  the  two  rectangles  into  which  the  figures 
are  divided  by  the  dotted  lines ;  that  is, 

_BHS      W 

°~~ir+ir 


'     I 

B 

*_ 

B 

G 

Q 

t 

Hh- 

1 

<-  b  >j 

I 

O 

T 

O 

FIG. 28 


The  position  of  the  centroidal  axis  GG  may  then  be  determined 
by  taking  moments  about  the  base  00.    That  is  to  say,  since  the 


FIRST  AND   SECOND  MOMENTS 


31 


total  area  is  A  =  BH  +  bh,  we  have,  by  the  principle  of  moments, 

TT  ~L 

x(BH  +  bJi)  =  BH  x  —  +  bh  X  -  ,  whence 

L  2 

•   BH*  +  IV 


Having  found  a?0,  the  /for  the  centroidal  axis  GG  is  determined  by 
the  relation  T    _  T        A* 

±G  —  •LO         -^^o* 


APPLICATIONS 

51.  A  uniform  rod  18  in.  long  weighs  8  lb.  and  has  weights  of  2  lb.,  3  lb.,  4  lb., 
and  5  lb.  strung  on  it  at  distances  of  6  in.  apart.  Find  the  point  at  which  the  rod 
will  balance. 

Solution.  Since  the  rod  is  uniform,  its  weight  may  be  assumed  to  be  concen- 
trated at  its  center.  If,  then,  x0  denotes  the  distance  of  the  center  of  gravity  from 
the  end  at  which  the  2-lb.  weight  is  hung,  by  taking  moments  about  this  end 

_  2  x  0+  3x6+8x9+4x12  +  5x18 

XQ    — 


2+3+8+4+5 


=  I<¥T 


52.    A 

t  =  ^  in. 


.4,J_r 
SP 


section  like  that  shown  in  Fig.  29  has  the  dimensions  6  =  3  in.,  d  =  5  in., 
Locate  its  center  of  gravity,  or  centroid. 

Solution.  To  locate  the  gravity  axis 


1  —  1,  take  moments  about  any  parallel 
line  as  a  base,  say  A  B.  Then,  dividing 
the  figure  into  two  rectangles,  since  the 
center  of  gravity  of  each  rectangle  is  at 
its  center,  we  have 


_ 


D 


Similarly,  to  determine  the  gravity  axis 
2  —  2,  by  taking  moments  about  CD  we 
have 


= 
4 


3X  i+4ix  i 


2 

FIG. 29 


53.   The  section  shown  in  Fig.  30  is 
made  up  of  two  10-in.  channels  30  Ib./ft. 
and  a  top  plate  9  in.  x  \  in.    Locate  its 
gravity  axes  and  determine  its  moment  of  inertia  with  respect  to  the  axis  1—1. 

Solution.   From  Table  IV  the  area  of  each  channel  is  8.82  in.2    To  determine 
the  gravity  axis  1  —  1,  take  moments  about  the  lower  edge  of  the  section.   Then 


2  x  8.82  +  9  x 


=  „  0? 


32 


RESISTANCE  OF  MATERIALS 


From  the  table,  the  moment  of  inertia  of  each  channel  with  respect  to  an  axis 
perpendicular  to  the  web  at  center  is  103.2  in.4,  and  the  distance  from  this  axis  to 


la 


jt 


ffl 


!3 

FIG. 30 


the  gravity  axis  of  the  entire  section  is  6.07  —  5  =  1.07  in.    Also,  the  moment  of 

inertia  of  the  top  plate  with  respect  to  its  gravity  axis  is  —  —     X  **'   =  —  in.4, 

12  12  32 

and  the  distance  of  this  axis  from  the 
gravity  axis  of  the  entire  section  is  4.18  in. 
Therefore 

Ii_i=  2 [103.2  +  8.82  x  (1.07)2]  +  [fa 

+  4.5  x  (4.18)2]  =305  in.4 

For  the  net  section  the  rivet  holes  must 
be  deducted  from  this  value.  Assuming 
two  | -in.  rivets,  the  amount  to  be  deducted 
is  approximately  24  in.4,  giving  for  the  net 
section  I^  =  281  in.* 

54.  In  problem  53  determine  the  mo- 
ment of  inertia  of   the  net  section  with 
respect  to  the  gravity  axis  2—2. 

55.  The  section   shown  in   Fig.  31   is 
made  up   of  four  angles  4  x  3  x  ^  in., 
with  the  longer  leg  horizontal,  and  a  web 
plate  12  x  |  in.,  with  f-in.  rivets.    Find 
the  moment  of  inertia  for  its  net  section 
with  respect  to  the  gravity  axis  1  —  1. 


lf 


FIG.  32 


56.  The  section  shown  in  Fig.  32  is  built  up  of  two  8-in.  channels  18.751b./ft. 
and  two  plates  9  x  f  in.  Find  the  moment  of  inertia  of  its  net  section  about  the 
gravity  axis  1—1,  deducting  the  area  of  four  J-in.  rivet  holes. 


FIKST  AND  SECOND  MOMENTS 


33 


57.  In  problem  56  find  the  moment  of  inertia  of  the  net  section  with  respect 
to  the  gravity  axis  2—2. 

58.  The  section  shown  in  Fig.  33  has  the  dimensions  6=10  in.,  d  =  4  in.,  t  =  1  in. 

Locate  the  gravity  axis  1  —  1. 

59.  In   problem   58    find 
the  moment  of  inertia  of  the 
section  with   respect  to  the 
gravity  axis  2  —  2. 

60.  The  section  shown  in 
Fig.  34  has  the   dimensions 


f  L  — 

UA 

.  

^__ 



t 

^_4__v 

* 

FIG. 

33 

tj  =  t2  =  ts  =  1  in.  Locate  the 
gravity  axis  1  —  1. 

61.  The  section  shown  in 
Fig.  32  is  composed  of  two  12-in. 
channels,  20.5  lb./ft.,  and  two 
^-in.  plates.  How  wide  must 
the  plates  be  in  order  that  the 
moments  of  inertia  of  the  section  shall  be  the  same  about  both  gravity  axes  ? 

62.  The    section    shown    in   Fig.  35 
has  the  dimensions  6  =  8  in.,  h  =  10  in., 
6'  =  5  in.,  h'  =  6  in.    Find  its  moments 
of  inertia  about  both  gravity  axes. 

63.  Two6-in.  channels  10. 5  lb./ft.  are 
connected  by  latticing.    How  far  apart 
should  they  be  placed,  back  to  back,  in 
order  that  the  moments  of  inertia  may 
be  the  same  about  both  gravity  axes  ? 

64.  A  hollow  cast-iron  column  is  6  in. 
external  diameter  and  1  in.  thick.   Find 
the  moment  of  inertia  of  its  cross  section 
with  respect  to  a  diameter. 

65.  The  section  shown  in  Fig.  31  is  made 
up  of  a  web  plate  9  x  $  in.  and  four  angles 
3  x  3  x  §  in.   Find  its  moment  of  inertia 

o 

with  respect  to  both  gravity  axes. 

66.  The  top  chord  of  a  bridge  truss 

has  a  section  like  that  shown  in  Fig.  36,  with  top  plate  20  x  f ",  two  web  plates 

each  18  x  f",  and  four  angles  3  x  3  x  f".  Find 
the  eccentricity  of  the  section ;  that  is,  the  dis- 
tance from  center  of  figure  to  gravity  axis  1  —  1. 
|  67.  In  problem  66  find  the  moment  of  iner- 

J  tia  of  the  net  section  with  respect  to  the  axis 

~~^ 1—1,  deducting  for  four  |-in.  rivets. 
I  68.  A  circular  table  rests  on  three  legs  placed 

I  at  the  edge  and  forming  an  equilateral  triangle. 

J,  Find  the  least  weight  which  will  upset  the  table 

FIG.  35  when  hung  from  its  edge. 


34 


RESISTANCE  OF  MATERIALS 


69.  A  horizontal  beam  20  ft.  long  and  weighing  120  Ib.  rests  on  two  supports 
10  ft.  apart.  A  load  of  75  Ib.  is  hung  at  one  end  of  the  beam  and  150  Ib.  at  the 
other  end.  How  must  the  beam  be  placed  so  that  the  pressure  on  the  supports 

may  be  equal  ? 

70.  Explain  how  a  clock  hand  on  a 
smooth  pivot  can  be  made  to  show  the  time 
by  means  of  clockwork  concealed  in  the 
hand  and  carrying  a  weight  around. 

71.   A  brick  wall  is  12  in.  thick  and  40  ft. 
high.  What  uniform  wind  pressure  will  cause 
-  -  -     it  to  tip  over  ?    Weight  of  ordinary  brick 
masonry  is  125  Ib./f t.3 

72.  A  masonry  dam  is  30  ft.  high,  6  ft. 
wide  at  top,    and  30  ft.  wide   at  bottom, 
with  upstream  face  vertical.   Assuming  the 
masonry  to  weigh  160  lb./ft.3,  compute  the 
moment  of  the  weight  of  the  dam  about 
the  toe  of  the  base. 

73.  In  problem  72  the  resultant  water 
pressure  for  a  vertical  strip  1  ft.  wide  is 
46,800  Ib.   and  is  applied  at  a  point  12ft. 

above  the  base  of  the  dam.    Determine  its  stability  against  overturning. 

74.  The  casting  for  a  gas-engine  piston  is  a  hollow  cylinder  of  uniform  thick- 
ness, with  one  end  closed.   The  external  diameter  is  5  in.,  length  over  all  6  in., 
thickness  of  cylinder  shell  Jin.,  thickness  of  end  l^in.    Find  the  distance  of  its 
center  of  gravity  from  the  closed  end. 

75.  A  cast-iron  pulley  weighs  50  Ib.  and  its  center  of  gravity  is  0.1  in.  out  of 
center.   To  balance  the  pulley,  a  hole  is  drilled  in  the  light  side,  6  in.  from  the 
center  of  the  pulley  and  in  line  with  its  center  of  gravity,  and  rilled  with  lead. 
How  much  iron  must  be  removed,  the  specific  gravity  of  lead  being  11.35  and  of 
iron  7.5  ? 


FIG.  36 


SECTION  III 

BENDING-MOMENT  AND  SHEAR  DIAGRAMS 

25.  Conditions  of  equilibrium.  In  order  that  any  structure  may 
be  in  equilibrium,  the  external  forces  acting  on  it  must  satisfy 
two  conditions : 

1.  The  sum  of  the  forces  acting  in  any  given  direction  must  be  zero. 

2.  The  sum  of  the  moments  of  the  forces  about  any  point  must  be  zero. 
If  force  and  moment  are  denoted  by  F  and  Jf,  respectively,  these 

conditions  are  expressed  more  briefly  in  the  form 

f2>=0; 

(28)  For  equilibrium  \  ~; 

If  the  forces  all  lie  in  one  plane,  the  condition  VJP  =  0  is  expressed 
more  conveniently  in  the  form 

(29)  ]>}  vertical  forces  =  O  ;       V  horizontal  forces  =  O. 

To  illustrate  the  application  of  these  conditions,  consider  a  simple 
beam  AB  of  length  Z,  supported  at  the  ends  and  bearing  a  single 
concentrated  load  P  at  a 


distance  d  from  one  end  A IP 

(Fig.  37).  Let  the  reac- 
tions of  the  supports  at  A 
and  B  be  denoted  by  Rv 
R  ,  and,  to  find  the  value 

2  FH;.  37 

01    these    reactions,    apply 

the  condition  VTJf  =  0  ;  that  is,  equate  to  zero  the  sum  of  the 
moments  of  all  the  external  forces  with  respect  to  any  convenient 
moment  center,  say  A.  Then 

Pd  -  RJ,  =  0, 

Pd 
whence  R^  =  —  • 

35 


36  RESISTANCE  OF  MATERIALS 

Now,  applying  the  condition  ^  vertical  forces  =  0,  we  have 

^  +  Rz  -  P  =  0, 

and  inserting  in  this  equation  the  value  just  found  for  R2  and  then 
solving  for  E^  the  result  is 


26.  Vertical  shear.  By  applying  the  conditions  V^  =  0,^Jf=0, 

•as  just  explained,  all  the  external  forces  acting  on  the  beam  may 
be  found.  The  beam  may  then  be  supposed  to  be  cut  in  two  at  any 
point  and  these  conditions  applied  to  the  portion  on  either  side  of 
the  section. 

In  general,  the  sum  of  the  external  forces  on  one  side  of  any 
arbitrary  cross  section  will  not  be  identically  zero.  If,  then,  the 
condition  of  equilibrium  V  F  —  0  is  satisfied  for  the  portion  of  the 
beam  on  one  side  of  the  section,  the  stress  in  the  material  at  this 
point  must  supply  a  force  equal  in  amount  and  opposite  in  direc- 
tion to  the  resultant  of  the  external  forces  on  one  side  of  this  point. 
This  resisting  force,  or  resultant  of  the  vertical  stresses  in  the  plane 
of  the  cross  section,  which  balances  the  external  forces  on  one  side 
of  the  section,  is  called  the  vertical  shear.  Therefore 

The  vertical  shear  on  any  cross  section  =  the  algebraic  sum  of  the 
external  vertical  forces  on  either  side  of  the  section. 

For  instance,  suppose  that  a  beam  10  ft.  long  bears  a  uniform 
load  of  300  lb./ft.,  and  it  is  required  to  find  the  vertical  shear  on 
a  section  4  ft.  from  the  left  support.  In  this  case  the  total  load 
on  the  beam  is  3000  lb.,  and,  since  the  load  is  uniform,  each  reac- 
tion is  1500  lb.  The  load  on  the  left  of  the  given  section  is  then 
4  x  300  =  1200  lb.,  and  therefore  -the  shear  at  the  section  is 
1500  -  1200  =  300  lb. 

27.  Bending  moment.    In  applying  the  condition  Vlf  =  0  to  the 
portion  of  a  beam  on  either  side  of  any  cross  section,  the  center  of 
moments  is  taken  at  the  centroid  of  the  section.    Since  the  position 
of  the  cross  section  is  arbitrary,  it  is  obvious  that  the  sum  of  the  mo- 
ments of  the  forces  on  one  side  of  the  section  about  its  centroid  will 
not  in  general  be  zero.   Therefore,  to  satisfy  the  condition  V  Jf  =  0, 
the  normal  stresses  in  the  beam  at  the  section  considered  must 


BENDING-MOMENT  AND  SHEAR  DIAGRAMS 


37 


P, 


supply  a  moment  which  balances  the  sum  of  the  moments  of  the 
external  forces  on  either  side  of  the  point.  This  resisting  moment 
in  the  beam  is  called  the  stress  couple  or  bending  moment,  and  is 
evidently  equal  to  the  resultant  external  moment  at  the  point  in 
question.  Consequently 

The  bending  moment  at  any  cross  section  of  a  beam  is  equal  to  the 
sum  of  the  moments  of  the  external  forces  on  one  side  of  this  point, 
about  the  centroid  of  the  section. 

For  example,  in  Fig.  38,  consider  a  cross  section  mn  at  an  arbitrary 
distance  x  from  the  left  support.  Then  for  the  portion  of  the  beam 
on  the  left  of  mn  the  mo- 
ment of  R1  about  the  cen- 
troid of  the  section  is  Rjc, 
and  the  moment  of  P^  about 
the  same  point  is  7J  (x  —  d^).  A 
Therefore  the  total  bend-  R\ 
ing  moment  at>  the  section 
mn  is 

M=Rx-Pl(x-d\         A 

R^T-  - x- 

As  another  example,  con- 
sider a  beam  of  length  I 
bearing  a  uniform  load  of  amount  w  per  unit  of  length.  Then  the 

total  load  on  the  beam  is  wl,  and  each  reaction  is  —  •    Therefore, 

2 

taking  a  section  at  a  distance  x  from  the  left  support  and  consider- 
ing only  the  forces  on  the  left  of  the  section,  the  total  bending 
moment  at  this  point  is 


\B 


Pi 


n 


- 


FIG.  38 


Wl  X        WX 

=  -—  •  x  —  wx  •  —  =  — 


From  this  relation  it  is  evident  that  M  =  0  when  x  =  0  or  x  =  Z,  and 
attains  its  maximum  value  when  x  =  -  ;  that  is,  the  bending  moment 

is  zero  at  each  end  of  the  beam  and  a  maximum  at  the  center. 

28.  Bending-moment  and  shear  diagrams.  Since  in  general  the 
bending  moment  and  shear  vary  from  point  to  point  along  a 
beam,  it  is  desirable  to  show  graphically  the  moment  and  shear  at 


38 


RESISTANCE  OF  MATERIALS 


each  point  of  the  beam.  This  may  be  done  by  means  of  a  bending- 
moment  diagram  and  a  shear  diagram,  obtained  by  plotting  the 
general  expressions  for  the  moment  and  shear,  such  as  those  given  in 
the  examples  in  the  preceding  paragraph.  Thus,  the  shear  diagram  is 
obtained  by  plotting  the  shear  at  any  arbitrary  section  mn  as  ordinate 
and  the  distance  x  of  this  section  from  a  fixed  origin  as  abscissa. 
Similarly,  the  moment  diagram  is  obtained  by  plotting  the  moment 
at  any  arbitrary  section  mn  as  ordinate  and  the  distance  x  as  abscissa. 

The  following  simple  applica- 
tions illustrate  the  method  of 
drawing  the  diagrams. 

1.  Simple  beam  bearing  a  single 
concentrated  load  P  at  its  center 
(Fig.  39).  From  symmetry,  the 
reactions  R^  and  R2  are  each  equal 

to  —  •    Let  mn  denote  any  section 

A 

of  the  beam  at  a  distance  x  from 
the  left  support,  and  consider  the 
portion  of  the  beam  on  the  left  of 
this  section.  Then  the  moment  at 

P 

9 


mn    s 


=  —  x]  and  the  shear 


=  —  ).    For  a  section  on  the 


FIG.  39  IS 

right  of  the  center  the  bending  moment  is  R^  (I  —  x)  and  the  shear 
is  R^.  Consequently,  the  bending  moment  varies  as  the  ordinates 
of  a  triangle,  being  zero  at  either  support  and  attaining  a  maximum 

PI 
value  of  —  at  the  center,  while  the  shear  is  constant  from  A  to  B, 

and  also  constant,  but  of  opposite  sign,  from  B  to  C. 

The  diagrams  in  Fig.  39  represent  these  variations  in  bending 
moment  and  shear  along  the  beam  under  the  assumed  loading. 
Consequently,  if  the  ordinates  vertically  beneath  B  are  laid  off  to 
scale  to  represent  the  bending  moment  and  shear  at  this  point,  the 
bending  moment  and  shear  at  any  other  point  D  of  the  beam  are 
found  at  once  from  the  diagram  by  drawing  the  ordinates  EF  and 
HK  vertically  beneath  Z>. 


BEKDING-MOMENT  AND  SHEAK  DIAGRAMS 


39 


2.  Beam  bearing  a  sin- 
gle concentrated  load  P 
at  a  distance  c  from  one 
support. 

The  reactions  in  this 
case  are 


and 


~ 


Hence,  the  bending  mo- 
ment  at  a  distance  x  from 
the  left  support  is 


provided  x  <  c,  and 
„  Pc(l-x) 

RtQ-x)  =  -  ~  —  *» 

0 

if  x  >  c.  If  x  =  c,  each  of 
these  moments  becomes 

PC  (l-c) 

~r 

and  consequently  the  bend- 
ing-moment  and  shear  dia- 
grams are  as  shown  in 
Fig.  40. 

3.  Beam  bearing  several 
separate  loads. 

In  this  case  the  bending- 
moment  diagram  may  be 
obtained  by  constructing 
the  diagrams  for  each  load 
separately  and  then  adding 
their  ordinates,  as  indicated 
in  Fig.  41. 


SHEAR 


FIG.  41 


40 


RESISTANCE  OF  MATERIALS 


4.  Beam  bearing  a  contin- 
uous uniform  load. 

Let  the  load  per  unit  of 
length  be  denoted  by  w. 
Then  the  total  load  on  the 
beam  is  ?rZ,  and  the  reac- 
tions are 


Hence,  at  a  distance  x  from 
the  left  support  the  bending 
moment  Mx  is 

wl  x 

Mx  =  —x-wx.- 


FIG.  42 


The  ben  ding-moment  diagram  is  therefore  a  parabola.   When  x  =  -< 


wl2      1.1.   ., 
Mx  =  — ,  which  is  its 

8 

maximum  value.  The 
bending-moment  and 
shear  diagrams  are 
therefore  as  repre- 
sented in  Fig.  42. 

5.  Beam  bearing  uni- 
form load  over  part  of 
the  span. 

Let  the  load  ex- 
tend over  a  distance 
c  and  be  of  amount 
w  per  unit  of  length. 
Then  the  total  load  is 
we.  The  reactions  of 
the  supports  are  the 
same  as  though  the 
load  were  concentrated 


FIG.  43 


BENDING-MOMENT  AND  SHEAR,  DIAGRAMS          41 

at  its  center  of  gravity  G.   Therefore,  if  d  denotes  the  distance  of 
G  from  the  left  support, 

,...  „  SI  J\ 

and          ft  = 


I 

Also,  the  bending-moment  diagrams  for  the  portions  AB  and  CD 
are  the  same  as  though  the  load  were  concentrated  at  G,  and  are 
therefore  the  straight  lines  A'H  and  D'K,  intersecting  in  the  point 
T  vertically  beneath  G  (Fig.  43). 

From  B  to  C  there  is  an  additional  bending  moment  due  to  the 
uniform  load  on  this  portion  of  the  beam.  Thus,  if  LMN  is  the 
parabolic  moment  diagram  for  a  beam  of  length  LN  or  c,  the  ordi- 
nates  to  the  line  HK  must  be  increased  by  those  to  the  parabola 
LMN,  giving  as  a  complete  moment  diagram  the  line  A'HJKD'. 

Analytically,  if  x  denotes  the  distance  of  any  section  from  the 
left  support,  the  equations  of  the  three  portions  A'H,  HJK,  and 
KD'  of  the  moment  diagram  are 

we  (I  —  d}  x  £  7      c 

MAB  =  MIX  =         *-y-    - '          ±or  ^  x  ^  d  -  - ; 

s.    n  , 

=  wc(l-d)x      "(*-'*  + 1 

2  I 

for 

V 

wed  (I  —  x)  c 

-y-A         for         d  +  ^xsl 

29.  Relation  between  shear  and  moment  diagrams.    Consider  a 
beam  bearing  any  number 
of    concentrated   loads   ^,  !PI 

P%,  •  •  «,  jfJJ,  at  distances  d^ 
c?2,  •  •  •,  dn  from  one  end  A 
(Fig.  44).  Then  the  mo-  _  ^^  ,  .  ,  ^ 


ment  M  at  any  section  ?rm, 

distant  x  from  the    origin  * 

4  is  F'°-44 


42  RESISTANCE  OF  MATERIALS 

where  the  summation  includes  only  the  loads  on  the  left  of  the 
section.  For  an  adjacent  section  distant  A#  from  mn,  that  is,  at  a 
distance  x  -f  Az  from  the  origin,  the  moment  is 


Let  &M  denote  the  difference  between  these  two  moments.    Then 
AJf  =  M'  -  M  =  Rx  -   T  PAz, 


--• 

But,  by  definition,  the  shear  S  at  the  given  section  mn  is 

S  =  S1-^/P.- 

Consequently, 


This  relation  also  holds  for  a  beam  uniformly  loaded.  Thus,  if 
w  denotes  the  uniform  load  per  foot  of  length,  and  I  is  the  span  in 
feet,  the  moment  in  this  case  at  any  section  distant  x  from  the  left 
support  is  wl 


and  at  a  section  distant  Ax  from  this  it  is 

w   x 


2 

Therefore  the  change  in  the  moment  is  now 

=  M'  -  M=kx-wx  .  Az 


If  Ao:  is  assumed  to  be  small,  its  square  may  be  neglected  in  com- 
parison with  the  other  terms.  In  this  case,  dropping  the  last  term, 
we  have 

AJf     wl 


Evidently  the  same  relation  holds  for  any  combination  of  uniform 
and  concentrated  loads.  The  general  fundamental  relation  between 
the  shear  and  moment  diagrams  is  therefore 


BENDING-MOMENT  AND  SHEAB,  DIAGRAMS          43 

Since  -  represents  the  rate  at  which  the  ordinate  to  the  moment 

A# 

diagram  is  changing,  this  relation  may  be  expressed  in  words  by 
saying  that 

The  rate  of  change  of  the  moment  is  equal  to  the  shear. 

From  this  result  important  properties  of  the  two  diagrams  may 
be  deduced,  as  explained  in  the  next  paragraph. 

30.  Properties  of  shear  and  moment  diagrams.  Consider  the 
highest  point  of  any  given  moment  diagram  —  for  instance,  of 
those  shown  in  Figs.  39-43. 

Since  the  moment  increases  up  to  this  point  and  decreases  after 
it  passes  it,  the  change  in  the  moment  AJf,  corresponding  to  an 
increase  A#  in  the  abscissa,  must  be  positive  on  one  side  of  the 
point  and  negative  on  the  other.  Since  Az  is  positive  in  both  cases, 

the  ratio  —  —  changes  sign  in  passing  the  point.   But  since  -  =  S, 

this  means  that  the  shear  changes  from  positive  to  negative  in  pass- 
ing the  given  point,  and  therefore  must  pass  through  zero  at  the 
point  in  question. 

The  same  reasoning  evidently  holds  for  the  lowest  point  of 
the  moment  diagram.  Therefore,  at  the  section  where  the  moment 
is  greatest  or  least  the  shear  is  either  zero  or  passes  through  zero 
in  passing  the  point. 

By  referring  to  the  diagrams  in  the  preceding  article  or  in 
Table  XIII  it  will  be  observed  that  this  is  true  in  each  case. 

If  the  moment  is  constant,  then  Alf=  0  and  consequently  S  =  0. 
That  is  to  say,  where  the  moment  is  constant  the  shear  is  zero. 

For  a  system  of  concentrated  loads  the  equations  for  moment 
and  shear,  as  shown  in  article  29,  are 


The  first  of  these  represents  an  inclined  straight  line,  and  the 
second  a  horizontal  straight  line.  Therefore,  for  concentrated  loads 
the  moment  diagram  is  a  broken  line  and  the  shear  diagram  is  a  series 
of  horizontal  lines  or  steps. 


44  RESISTANCE  OF  MATERIALS 

For  a  uniform  load  the  expressions  for  moment  and  shear,  as 

shown  in  article  29,  are 

wl        wx2 


0      wl 

S  =  ~-wx. 

The  first  of  these  equations  evidently  represents  a  parabola,  and  the 
second  an  inclined  straight  line  of  slope  =  w. 

From  these  results  it  follows  that  for  any  combination  of  uniform 
and  concentrated  loads  the  moment  diagram  is  a  connected  series  of 
parabolic  arcs,  and  the  shear  diagram  is  a  succession  of  inclined  lines 
or  sloping  steps. 

Since  S&x  represents  an  elementary  vertical  strip  of  the  shear 
diagram,  the  area  subtended  by  the  shear  diagram  between  any  two 
given  points  is  V$A:r.  Making  use  of  the  relation  AjM"=£A#,  and 
summing  between  two  points  ^  and  P^  we  have 


where  M^  and  M2  denote  the  moments  at  the  two  points  in  question. 
Hence  the  difference  between  the  moments  at  any  two  given  points  is 
equal  to  the  area  of  the  shear  diagram  between  these  points. 

At  the  ends  of  a  simple  beam  the  moment  is  always  zero.  There- 
fore, by  the  theorem  just  proved,  for  a  simple  beam  the  area  of  the 
shear  diagram  from  one  end  to  any  point  is  equal  to  the  moment  at 
this  point. 

31.  General  directions  for  sketching  diagrams.  To  economize  time 
and  effort  it  is  important  to  follow  a  definite  program  in  drawing 
the  diagrams  and  determining  the  expressions  for  shear  and  moment. 
The  following  outline  of  procedure  for  either  cantilever  beams  or 
simple  beams  resting  on  two  supports  is  therefore  suggested. 

1.  Find  each  reaction  by  summing  the  moments  of  all  the  ex- 
ternal forces  about  a  point  on  the  opposite  reaction  as  moment 
center.    Check  this  calculation  by  noting  that  the  sum  of  the  reac- 
tions must  equal  the  sum  of  the  loads. 

2.  Note  that  the  expressions  for  moment  and  shear  both  change 
whenever  a  concentrated  load  is  passed.    Consequently,  there  will 


BENDING-MOMENT  AND  SHEAR  DIAGRAMS         45 


inear-  foo 


:ale: 


Idiv 


2100  II A  fc> 


Scae 


200 


Shea 


lagram 


Dlajr 


Ibs. 


FIG.  45 

be  as  many  different  segments  of  the  moment  and  shear  diagrams 
as  there  are  segments  of  the  beam  between  concentrations. 

3.  For  a  simple  beam,  take  the  origin  at  the  left  end  of  the  beam. 
For  a  cantilever  beam,  take  the  origin  at  the  free,  or  unsupported,  end 
of  the  beam.  Keep  the  origin  at  this  point  throughout  the  calculations. 


46  RESISTANCE  OF  MATERIALS 

4.  Take  a  section  between  the  origin  and  the  first  concentration,  let 
x  denote  the  distance  of  this  section  from  the  origin,  and  find  the  gen- 
eral expressions  for  the  moment  and  shear  at  this  section  in  terms  of  x. 

5.  Proceed  in  the  same  way  for  a  section  between  each  pair  of 
consecutive  concentrations. 

6.  Plot  these   equations,  checking  the  work  by  means  of  the 
general  relations  stated  in  the  preceding  article. 

7.  Plot  the  shear  diagram  first.    In  plotting  this  diagram  it  is 
convenient  to  follow  the  direction  in  which  the  forces  act.    Thus,  in 
Fig.  45  the  shear  at  the  left  end  is  equal  to  the  reaction  and  may 
be  laid  off  in  the  same  direction,  that  is,  upwards.    Proceeding  to 
the  right,  drop  the  shear  diagram  by  an  amount  equal  to  each  load 
as  it  is  met,  until  the  reaction  at  the  right  end  is  reached,  which  will 
bring  the  shear  diagram  back  to  the  base  line.    By  following  this 
method  the  shear  diagram  will  always  begin  and  end  on  the  base 
line,  which  serves  as  a  check  on  the  work. 

8.  Note  that  as  long  as  the  shear  diagram  lies  above  the  base  line  the 
shear  is  positive  and  therefore  A M  is  also  positive  ;  that  is  to  say,  the 
moment  is  increasing.     Where  the  shear  diagram  crosses  the  axis, 
the  moment  diagram  must  attain  its  highest  or  lowest  point.    When 
the  shear  diagram  lies  below  the  base  line,  the  moment  is  decreasing. 

9.  Compute  numerical  values  of  the  moment  and  shear  at  the 
critical  points  of  the  diagrams,  and  indicate  these  numerical  values 
on  the  diagrams. 

A  sample  set  of  diagrams  as  they  should  be  drawn  by  the  student 
is  shown  in  Fig.  45. 

APPLICATIONS 

76.  A  beam  16  ft.  long  is  supported  at  the  left  end  and  at  a  point  4  ft.  from  the 
right  end,  and  carries  a  uniform  load  of  200  Ib./ft.  over  its  entire  length  and  a 
concentrated  load  of  1  ton  at  a  point  4  ft.  from  the  left  end.  Sketch  the  shear 
and  moment  diagrams  and  note  the  maximum  shear  and  maximum  moment. 

Solution.   On  cross-section  paper  indicate  the  loading  as  shown  in  Fig.  45. 

To  find  either  reaction,  take  moments  about  the  other  point  of  support.  Thus, 
for  the  left  reaction  Rl  we  have 

Rl  •  12  -  3200  .  4  +  2000  -8  =  0,     whence     R^  =  2400  Ib. 
Similarly,  for  R2,      E2  .  12  -  3200  .  8  -  2000  -4  =  0,     whence     R2  =  2800  Ib. 

As  a  check  on  the  correctness  of  these  results,  sum  of  loads  is  3200  +  2000  =  5200, 
and  sum  of  reactions  is  2400  +  2800  =  5200. 


BENDING-MOMENT  AND  SHEAR  DIAGRAMS          47 

To  obtain  the  shear  diagram,  start  at  the  left  end  and  lay  off  the  reaction  of 
2400  Ib.  upward.  Since  the  load  is  200  lb./ft.,  at  4  ft.  from  the  left  end  the  shear 
will  be  2400—  4  x  200  =  1600  Ib.  As  we  pass  this  point  the  concentrated  load  of  1  ton 
will  cause  the  shear  to  drop  to  1600  —  2000  =  —  400  Ib.  The  shear  then  continues 
to  drop  200  lb./ft.,  until  at  the  right  support  it  becomes  —  400^—  8  x  200  =  —  2000  Ib. 
As  this  point  is  passed,  the  reaction,  which  is  equivalent  to  a  concentrated  load  of 
2800  Ib.  upward,  causes  the  shear  to  change  suddenly  to  —  2000  +  2800  =  800  Ib. 
It  then  gradually  drops  again  and  becomes  zero  at  the  end  of  the  beam. 

On  account  of  the  uniform  load  the  moment  diagram  will  be  segments  of  para- 
bolas. To  plot  these  parabolas  the  values  of  the  moment  at  a  number  of  points 
along  the  beam  may  be  calculated.  Thus,  at  points  2,  4,  10, 12,  and  14  ft.  from  the 
left  end  the  moments  are 

M2   =  2400  •  2  -  400  :  I  =  4400  f t.-lb.  ^ 

M4  =  2400  .  4  -  800  •  2  =  8000  f  t.-lb. 

M1Q  =  2400  .  10  -  2000  •  5  -  2000-  6  =  2000  f  t.-lb. 

M12  =  2400  •  12  -  2400  ,  6  -  2000  -  8  =  -  1600  f  t.-lb. 

Mu  =  2400  •  14  -  2800  .  7  -  2000  .  10  =  -  400  ft.-lb. 

The  maximum  moment  is  evidently  at  the  1-ton  load,  and  the  maximum  shear  at 
the  left  support. 

77.  A  simple  beam  10  ft.  long  is  supported  at  the  ends  and  carries  a  load  of 
800  Ib.  at  a  point  4  ft.  from  the  left  end.    Draw  the  shear  and  moment  diagrams.** 

78.  A  simple  beam  20  ft.  long,  supported  at  the  ends,  carries  a  uniform  load  of 
50  lb./ft.  and  a  'concentrated  load  of  600  Ib.  at  5  ft.  from  the  right  end.    Draw 
the  shear  and  moment  diagrams. 

79.  A  simple  beam  of  15  ft.  span  is  supported  at  the  ends  and  carries  a  uniform 
load  of  100  lb./ft.  and  concentrated  loads  of  500  Ib.  at  4  ft.  from  the  left  end  and 
1000  Ib.  at  8  ft.  from  the  left  end.    Plot  the  shear  and  moment  diagrams. 

80.  A  simple  beam  of  16ft.  span  carries  concentrated  loads  of  200  lb.,.400  Ib., 
and  100  Ib.  at  distances  of  4  ft.,  8  ft.,  and  12  ft.,  respectively,  from  the  left  support. 
Neglecting  the  weight  of  the  beam  itself,  sketch  the  shear  and  moment  diagrams. 

81.  A  simple  beam  of  9  ft.  span  carries  a  total  uniform  load  of  400  Ib.  over  the 
middle  third  of  the  span.    Neglecting  the  weight  of  the  beam,  draw  the  shear  and 
moment  diagrams  for  this  loading. 

82.  The  total  load  on  a  car  axle  is  8  tons,  equally  divided  between  the  two 
.wheels.  .Distance  between  centers  of  wheels  is  4^  ft.,  and  distance  between  centers 

o|  journals  is  5^  ft.  Draw  the  shear  and  moment  diagrams  for  the  axle  so  loaded. 
A"83.  Draw  the  shear  and  moment  diagrams  for  a  simple  beam  10  ft.  long,  bear- 
ing ale*al  uniform  load  of  100  lb./ft.  and  concentrated  loads  of  1  ton  at  4  ft.  from 
the  left  end  and  2  tons  at  3  ft.  from  the  right  end. 

84.  A  [beam  12  ft.  long  is  supported  at  the  ends  and  carries  loads  of  4000  Ib. 
and  1000  lb!  at  2  ft.  and  4  ft.,  respectively,  from  the  left  end.    No  uniform  load. 
Sketch  the  shear  and  moment  diagrams. 

85.  A  beam  20  ft.  long,  supported  at  the  ends,  bears  a  uniform  load  of  100  lb./ft. 
extending  from  the  left  end  to  the  center,  and  a  concentrated  load  of  1000  Ib.  at 
5  ft.  from  the  right  end.    Plot  the  shear  and  moment  diagrams. 

86.  A  beam  16  ft.  long,  supported  at  the  ends,  carries  a  uniform  load  of  200  lb./ft. 
extending  10  ft.  from  the  left  end,  and  concentrated  loads  of  1  ton  and  \  ton  at  8  ft. 
and  12  ft.,  respectively,  from  the  left  end,   Draw  the  shear  and  moment  diagrams. 


48  RESISTANCE  OF  MATERIALS 

87.  A  simple  beam  of  8  ft.  span  bears  a  distributed  load  which  varies  linearly 
from  zero  at  one  end  to  a  maximum  at  the  other.   The  total  load  on  the  beam  is 
1200  Ib.   Plot  the  shear  and  moment  diagrams. 

88.  A  cantilever  beam  extends  9  ft.  from  a  wall  and  bears  a  uniform  load  of 
60  Ib./ft.  and  a  concentrated  load  of  175  Ib.  at  the  free  end.   Draw  the  shear  and 
moment  diagrams. 

89.  A  cantilever  beam  projects  6  ft.  and  supports  a  uniform  load  of  100  Ib./f t. 
and  concentrated  loads  of  90  Ib.  and  120  Ib.  at  points  2ft.  and  4ft.,  respectively, 
from  the  free  end.   Draw  the  shear  and  moment  diagrams. 

90.  A  cantilever  beam  projects  10  ft.  and  carries  a  concentrated  load  of  100  Ib. 
at  the  free  end  and  also  concentrated  loads  of  90  Ib.  and  60  Ib.  at  3ft.  and  5ft., 
respectively,  from  the  free  end.    Sketch  the  shear  and  moment  diagrams. 

91.  A  cantilever  beam  projects  6  ft.  from  its  support  and  bears  a  concentrated 
load  of  50  Ib.  upward  at  the  free  end  and  50  Ib.  downward  at  2  ft.  from  the  free 
end.    Draw  the  shear  and  moment  diagrams,  neglecting  the  weight  of  the  beam. 

92.  A  cantilever  beam  projects  8  ft.  from  its  support  and  bears  a  distributed 
load  which  varies  linearly  from  zero  at  the  free  end  to  a  maximum  at  the  fixed 
end.   The  total  load  is  ^  ton.    Draw  the  shear  and  moment  diagrams. 

93.  Sketch  the  shear  and  moment  diagrams  for  a  cantilever  12  ft.  long,  carrying 
a  total  uniform  load  of  50  Ib./ft.  and  concentrated  loads  of  200  Ib.,  150  Ib.,  and 
400  Ib.  at  distances  of  2  ft.,  4  ft.,  and  7  ft.,  respectively,  from  the  fixed  end. 

94.  An  overhanging  beam  of  length  30  ft.  carries  concentrated  loads  of  1  ton  at 
the  left  end,  1.5  tons  at  the  center,  and  2  tons  at  the  right  end,  and  rests  on  two 
supports,  one  4  ft.  from  the  left  end  and  the  other  6  ft.  from  the  right  end.    Draw 
the  shear  and  moment  diagrams. 

95.  An  overhanging  beam  20  ft.  in  length  bears  a  uniform  load  of  100  Ib./ft. 
and  rests  on  two  supports  10  ft.  apart  and  5  ft.  from  the  ends  of  the  beam. 
Sketch  the  shear  and  moment  diagrams. 

96.  An  overhanging  beam  25  ft.  in  length  carries  a  uniform  load  of  200  Ib./ft. 
over  its  entire  length,  and  rests  on  two  supports,  one  at  the  right  end  and  the 
other  at  10  ft.  from  the  left  end.    Plot  the  shear  and  moment  diagrams. 

97.  An  overhanging  beam  40  ft.  in  length  is  supported  at  points  4  ft.  from  the 
left  end  and  8  ft.  from  the  right  end.    It  carries  concentrated  loads  of  4  tons  at 
the  left  end,  3  tons  at  6  ft.  from  the  left  end,  2  tons  at  14  ft.  from  the  left  end,  and 
1  ton  at  the  right  end.    Draw  the  shear  and  moment  diagrams. 

98.  Draw  the  shear  and  moment  diagrams  for  an  overhanging  beam  18  ft.  in 
length,  supported  at  points  4  ft.  from  each  end,  and  carrying  a  uniform  load  of 
50  Ib./ft.  over  its  entire  length  and  a  concentrated  load  of  800  Ib.  at  the  middle. 

99.  Draw  the  shear  and  moment  diagrams  for  an  overhanging  beam  20  ft.  in 
length,  supported  at  points  3  ft.  from  the  left  end  and  5  ft.  from  the  right  end, 
which  carries  a  uniform  load  of  80  Ib./ft.  between  the  supports  and  concentrated 
loads  of  600  Ib.  at  each  end. 

100.  Draw  the  shear  and  moment  diagrams  for  an  overhanging  beam  16  ft.  in 
length,  supported  at  points  2  ft.  from  the  left  end  and  4  ft.  from  the  right  end, 
which  carries  a  load  of  200  Ib./ft.  distributed  uniformly  over  12  ft.  from  the  left 
end,  and  a  concentrated  load  of  1600  Ib.  at  the  right  end. 


SECTION  IV 


STRENGTH  OF  BEAMS 

32.  Nature  of  bending  stress.  For  a  horizontal  beam  carrying  a 
set  of  vertical  loads  the  method  just  explained  for  drawing  the 
moment  and  shear  diagrams  is  to  combine  the  forces  on  one  side 
of  any  cross  section  into  a  single  force,  arid  the  moments  of  these 
forces  about  the  centroid  of  the  section  into  a  single  moment.  For 
equilibrium  the  stresses  in  the  beam  at  the  given  section  must 
therefore  also  reduce  to  a 
single  force  and  moment, 
called  the  shear  and  bend- 
ing moment,  respectively, 
equal  in  amount  and  op- 
posite in  direction  to  the 
external  resultant  force 
and  moment. 

By  considering  a  few 
simple  cases  the  nature 
of  the  shearing  and  bend- 
ing stresses  will  be  ap- 
parent. Thus,  in  Fig.  46, 
suppose  that  a  small  ver- 
tical slice  is  cut  out  of 
the  beam,  as  shown ;  then 
there  will  evidently  be  a  tendency  for  the  top  of  the  cut  to  close 
up  and  for  the  lower  side  to  spread  apart.  This  might  be  prevented 
by  placing  a  small  block  in  the  upper  edge  of  the  cut  and  connect- 
ing the  lower  edges  with  a  link.  Supposing  this  to  be  done,  there 
will,  in  general,  still  be  a  tendency  for  the  part  on  one  side  of  the 
cut  to  slide  up  or  down  past  the  part  on  the  other  side.  To  pre- 
vent this  vertical  motion,  it  would  be  necessary  to  introduce  a 
vertical  support,  as  shown  in  the  lower  diagram  of  Fig.  46. 

49 


^  Compression 
O O  Tension 


\J 


Vertical  Shear 


FIG.  46 


50  RESISTANCE  OF  MATERIALS 

From  this  illustration  it  is  evident  that  the  resisting  stress  in  a 
beam  required  to  equilibrate  any  system  of  external  forces  is  of 
two  kinds : 

1.  A  compressive  stress  on  one  side,  normal  (that  is,  perpen- 
dicular) to  the  plane  of  the  cross  section. 

A  tensile  stress  on  the  opposite  side,  also  normal  to  the  plane  of 
the  cross  section. 

2.  A  vertical  shearing  stress  in  the  plane  of  the  cross  section. 
33.  Distribution  of  stress.    The  effect  of  the  external  bending 

moment  on  a  beam  originally  straight  is  to  cause  its  axis  to  become 
^  bent  into  a  curve,  called  the  elastic  curve. 

fi   j  \  Considering  the   beam  to   be    composed  of 

/   \/      \  single  fibers  parallel  to  its  axis,  it  is  found 

/     /\       \  by  experiment  that  when  a  beam  is  bent,  the 

fibers  on  one  side  are  lengthened  and  those 
on  the  other  side  are  shortened.  Between 
these  there  must  evidently  be  a  layer  of  fibers 
which  are  neither  lengthened  nor  shortened, 
but  retain  their  original  length.  The  line  in 
which  this  unstrained  layer  of  fibers  inter- 
sects any  cross  section  is  called  the  neutral 
axis  (Fig.  47). 

It  is  also  found  by  experiment  that  a  cross 

section  of  the  beam  which  was  plane  before  flexure  (bending)  is 
plane  after  flexure.  This  is  known  as  Bernoulli's  assumption.*  As 
a  consequence  of  Bernoulli's  assumption  it  is  evident  from  Fig.  47 
that  the  lengthening  or  shortening  of  any  longitudinal  fiber  is  pro- 
portional to  its  distance  from  the  neutral  axis.  But  by  Hooke's 
law  the  stress  is  proportional  to  the  deformation  produced.  There- 
fore the  normal  stress  at  any  point  in  the  cross  section  is  likewise 
proportional  to  the  distance  of  this  point  from  the  neutral  axis.  If, 
then,  the  normal  stresses  are  plotted  for  every  point  of  any  vertical 
strip  MN  (Fig.  48),  their  ends  will  all  lie  in  a  straight  line.  This 
distribution  of  stress  is  therefore  called  the  straight-line  law. 

*  St.  Venant  has  shown  that  Bernoulli's  assumption  is  rigorously  true  only  for  certain 
forms  of  cross  section.  If  the  bending  is  slight,  however,  as  is  the  case  in  all  structural 
work,  no  appreciable  error  is  introduced  by  assuming  it  to  be  true  whatever  the  form 
of  cross  section. 


STRENGTH  OF  BEAMS  51 

Since  the  normal,  or  bending,  stresses  are  the  only  horizontal  forces 
acting  on  the  portion  of  the  beam  considered,  in  order  to  satisfy 
the  condition  of  equilibrium  horizontal  forces  =  0  we  must  have 


Resultant  tensile  stress  =  resultant  compressive  stress. 

Therefore,  since  the  tensile  and  compressive  stresses  act  in  opposite 

directions  (that  is,  are  of  opposite  sign),  the  algebraic  sum  of  all  the 

normal  stresses  acting  on  the   section  must 

be  zero.    Thus,  if  AA  denotes  an  element  of 

area  of  the  section,  and  p  the  intensity  of  the 

normal  stress  acting  on  it,  the  total  stress  on 

this  area  is  jt?A^4,  and  consequently 

=  0.  FIG.  48 

Now,  if  the  normal  stress  at  a  variable  distance  y  from  the  neutral 
axis  is  denoted  by  JK>,  and  that  at  some  fixed  distance  y'  is  denoted 

by  p',  then,  from  the  straight-line  law,  —,  =  —,->  or 

p'      y' 


Inserting  this  value  of  p  in  the  above  condition  of  equilibrium, 
it  becomes  , 


Therefore,  since  p'  and  y1  are  definite  quantities  different  from  zero, 
we  have  VyA^4  =  0.  But,  from  article  13,  the  distance  of  the  centroid 
from  the  neutral  axis  is  given  by 


and  if  ^yt±A  =  0,  then  also  «/0  =  0.  Therefore  the  neutral  axis 
passes  through  the  centroid  of  the  cross  section  ;  that  is,  the  neu- 
tral axis  coincides  with  the  horizontal  centroidal  axis. 

34.  Fundamental  formula  for  beams.  For  equilibrium  the  result- 
ant moment  of  the  normal  stresses  acting  on  any  cross  section  must 
be  equal  to  the  resultant  moment  of  the  external  forces  on  one 
side  of  the  section,  taken  with  respect  to  the  neutral  axis  of  the 


52  RESISTANCE.  OF  MATERIALS 

section.  Now,  if  A^4  denotes  an  element  of  area  of  the  cross  section, 
and  pf  the  intensity  of  the  normal  stress  acting  on  it,  the  total  stress 
on  this  area  is  p'&A.  If,  then,  y  is  the  distance  of  this  stress,  or 
internal  force,  from  the  neutral  axis  of  the  section,  and  M  denotes 
the  resultant  moment  of  the  external  forces  about  this  axis,  for 
equilibrium 


Now  let  p  denote  the  stress  on  the  extreme  fiber  and  e  the  distance 
of  this  fiber  from  the  neutral  axis.   Then,  by  the  straight-line  law, 

£.'=£, 

y     e' 

and,  inserting  this  value  of  p'  in  the  above  equation,  it  becomes 


The  quantity  V?/2A^4,  however,  is  the  moment  of  inertia,  /,  of  the 
cross  section  (article  20).  Therefore 

(33)  M  =  ££• 

e 

The  right  member  of  this  equation,  ^—  ,  is  the  resultant  internal 

e 

stress  couple,  and  is  called  the  moment  of  resistance  of  the  beam. 
Since  e  denotes  the  distance  of  the  extreme  fiber  of  the  beam 

from  the  neutral  axis,  the  ratio  -  is  also  a  function  of  the  shape 
and  size  of  the  cross  section,  and  is  therefore  called  the  section 

modulus.  Let  this  section  modulus  be  denoted  by  Z.  Then  Z  ==  -  , 
and  the  fundamental  formula  becomes 

(34)  M  =  pZ. 

Since  this  is  an  equality  between  the  resultant  external  moment  M 
and  the  product  of  the  working  stress  p  by  the  section  modulus  Z, 
it  expresses  the  fact  that  the  strength  of  a  beam  depends  jointly  on 
the  shape  and  size  of  the  cross  section  and  the  allowable  stress  for 
the  material. 

35.  Calculation  and  design  of  beams.  For  a  beam  of  given  size 
and  loading  the  maximum  external  moment  M,  acting  at  any  point 
along  the  beam,  is  first  determined  by  the  methods  explained  in 


STRENGTH  OF  BEAMS  53 

Section  III.  The  section  modulus  Z  is  then  calculated  from  the 
given  dimensions.  For  ordinary  rolled  shapes  of  structural  steel 
the  section  moduli  are  given  in  Tables  III-  VII.  The  stress  in  the 
extreme  fiber  (or  skin  stress,  as  it  is  called)  is  then  found  by  substi- 
tuting these  numerical  values  of  M  and  Z  in  the  equation  . 

M 
P  =  ~Z' 

By  comparing  this  calculated  value  of  p  with  the  allowable  unit 
stress  for  the  material,  it  is  determined  whether  or  not  the  beam 
is  safe. 

For  a  beam  of  given  size  and  shape  the  maximum  external 
moment  it  can  carry  safely  is  found  by  calculating  its  moment  of 
resistance.  Thus,  if  p  denotes  the  allowable,  or  working,  stress  for 
the  material  in  lb./in.2,  and  the  section  modulus  Z  is  calculated 
from  the  given  dimensions  of  the  cross  section,  the  maximum 
external  moment  M  which  this  beam  can  carry  with  safety  is  found 
by  inserting  these  numerical  values  in  the  equation 

M=pZ. 

In  designing  a  beam  to  carry  a  given  loading,  the  maximum 
external  moment  M  due  to  this  loading  is  first  calculated.  Then, 
for  any  specified  unit  working  stress  p,  the  required  section  modulus 
is  found  from  the  relation  ,, 

"  P' 

This  section  modulus  Z  may  then  be  looked  up  in  Tables  III-  VI, 
thus  determining  the  exact  dimensions  of  the  beam. 

APPLICATIONS 

101.  Find  the  safe  moment  of  resistance  for  an  oak  beam  8  in.  deep  and 
4  in.  wide. 

Solution.  In  this  case  1=  170.7  in.4  and  e  =  4  in.  Therefore  the  section  modu- 
xus  is 


From  Table  I  the  safe  stress  for  timber  may  be  assumed  as  p  —  1000  lb./in, 
Consequently,  the  moment  of  resistance  for  this  beam  is 


54 


RESISTANCE  OF  MATERIALS 


102.  In  an  inclined  railway  the  angle  of  inclination  with  the  horizontal  is 
30°.    The  stringers  are  10  ft.  6  in.  apart,  inside  measurement,  and  the  rails  are 

placed  1  ft.  inside  the  stringers.  The 
ties  are  8  in.  deep  and  6  in.  wide,  and 
the  maximum  load  transmitted  by  each 
rail  to  one  tie  is  10  tons.  Calculate  the 
maximum  normal  stress  in  the  tie. 

Solution.  The  bending  moment  is 
the  same  for  all  points  of  the  tie  be- 
tween the  rails,  and  is  20,000  ft.-lb. 
The  components  of  the  moment  with 
respect  to  the  axes  of  the  section 


FIG.  49 


(Fig.  49)  are  Mz  =  240,000 


and  My  =  240,000  (^)  in.-lb.,  and  the  section  moduli  with  respect  to  these  axes  are 
Zz  =  64  in.3  and  Zy  =  48  in.3  Therefore  the  maximum  normal  stress  is 


Pmax  = 


240,000    --)       240,000  - 


(54 


48 


=  57441b./in.2 


103.  A  rectangular  cantilever  projects  a  distance  I  from  a  brick  wall  and 
bears  a  single  concentrated  load  P  at  its  end.  How  far  must  the  inner  end 
of  the  cantilever  be  em- 
bedded in  the  wall  in 
order  that  the  pressure 
between  this  end  and 
the  wall  shall  not  exceed 
the  crushing  strength 
of  the  brick? 

Solution.  Let  b  denote 
the  width  of  the  beam 
and  x  the  distance  it 
extends  into  the  wall. 
For  equilibrium  the  re- 
action between  the  beam 
and  the  wall  must  con- 
sist of  a  vertical  force 
and  a  moment.  If  pa 
denotes  the  intensity  of 

the  vertical  stress,  and  it  is  assumed  to  be  uniformly  distributed  over  the  area  6x, 

p 
then  pahx  =  P  ;  whence  pa  =  —  (see  Fig.  50,  a). 

Similarly,  let  pb  denote  the  maximum  intensity  of  the  stress  forming  the  stress 
couple.  Then,  taking  moments  about  the  center  C  of  the  portion  AB,  since  the 
stress  forming  the  couple  is  also  distributed  over  the  area  &x,  we  have 


12 


and 


STRENGTH  OF  BEAMS 

Therefore,  substituting  in  the  formula  p  —  — ,  we  have 


55 


bx2 


12 


6P 


Consequently, 

whence 

and 


=  Pb 


(-1) 


,£. 

ta2  fee ' 


2P/      3z\ 

=  —  (2+—    , 

bx  \        x/ 


2P 


As  a  numerical  example,  let  I  =  5ft.,  P  =  200  lb.,  6  =  4  in.,  and  p  =  600  lb./in.2 
(for  ordinary  brickwork).  Then,  solving  the  above  equation  by  the  formula  for 
quadratics, 


X  = 


2P± 


bp 


whence,  by  substituting  the  above 
numerical  values, 

x  =  5.6  in. 

104.  Find  the  required  dimen- 
sions for  the  arms  of  a  cast-iron 
pulley  of  external  diameter  D  for 
a  tension  in  the  belt  of  T^  on  the 
tight  side  and  T2  on  the  slack  side. 
Arms  assumed  to  be  elliptical  in  cross  section,  of  dimensions  h  =  2b  (Fig.  51). 

Solution.  According  to  Bach  the  load  may  be  assumed  to  be  carried  by  one 
third  of  the  spokes,  and  the  working  stress  taken  as  4500  lb./in.'2  Let  n  denote 
the  number  of  spokes.  Then  the  maximum  moment  on  one  spoke  is  approximately 


FIG. 51 


The  moment  of  inertia  of  an  ellipse  about  its  minor  axis  is ,  and  its  section 

h2  M 

modulus  is  Z  =  —  Therefore,  substituting  these  values  in  the  formula  p  =  — , 

,                       16  ^ 
we  have 


whence 


h  = 


56 


RESISTANCE  OF  MATERIALS 


105.  Derive  a  formula  for  the  pitch  of  a  cast-iron  gear  to  carry  safely  a  driving 
force  F. 

Solution.  Circular  pitch  is  defined  as  the  distance  between  corresponding  points 
on  two  successive  teeth,  measured  along  the  pitch  circle.  Let  P  denote  the  circular 

pitch  for  the  case  in  question  (Fig.  52). 
Then,  if  h  denotes  the  depth  of  the 
tooth,  b  its  breadth,  and  t  its  thickness 
at  the  root,  the  relative  proportions 


FlG'  52 


ordinarily  used  are 
h=.7P,       £  =  .5P 
BC  =  .47  P, 


i 


6  =  2Pto3P, 
AB  =  .53  P, 


Height  above  pitch  circle  (called  addendum)  =  .3  P, 
Depth  within  pitch  circle  =  .4  P. 

The  driving  force  -F  is  ordinarily  applied  tangent  to  the  pitch  circle.  Assume,  how- 
ever, that  by  reason  of  the  gear  being  worn,  or  from  some  other  cause,  it  reaches  the 
tip  of  the  tooth,  as  shown  in  the  figure.  Then,  considering  the  tooth  as  a  cantilever 
beam,  the  maximum  moment  is 

M=Fh, 

and  its  section  modulus  at  the  root  is 


6 
Therefore,  assuming  a  working  stress  for  cast  iron  of  p  =  4500  lb./in.2,  we  have 


4500 "—  =  Fh,  i< 8'L J  . 

L ! . — i — L. 

and,  inserting  the  values  I — . . — . . — I— 1__ — 


this  becomes 


k" 


106.  Find  the  moment  of  resist-  >   J'  ^ ^  12" 

ance  for  the  section  given  in  problem 

53,  assuming  the  working  stress  for  i 

structural  steel  as  16,000  lb./in.2 

107.  Find   the    section   modulus 
and  moment  of  resistance  for  the 
section  given  in  problem  55.  p 

108.  Find  the   section   modulus     | I t 

and  moment  of   resistance   for  the  FIG.  53 

section  given  in  problem  56. 

109.  Find  the  moment  of  resistance  of  a  circular  cast  iron  beam  6  in.  in  diameter. 

110.  Find  the  moment  of  resistance  of  a  24-in.  steel  I-beam  weighing  80  Ib./ft. 

111.  Compare  the  moments  of  resistance  of  a  rectangular  beam  8  in.  x  14  in. 
in  cross  section,  when  placed  on  edge  and  when  placed  on  its  side. 


STRENGTH  OF  BEAMS 


57 


112.  Find  the  section  moduli  for  the  sections  given  in  problems  58,  60,  and  62. 

113.  Design  a  steel  I-beam,  10  ft.  long,  to  bear  a  total  uniform  load  of  1500  Ib./f t. 
including  its  own  weight. 


114.  A  built  beam  is  to  be  composed  of  two  steel  channels  placed -on  edge  and 
connected  by  latticing.    What  must  be  the  size  of  the  channels  if  the  beam  is 
to  be  18  ft.  long  and  bear  a  load  of  10  tons  at  its 

center,  for  a  working  stress  of  16,000  lb./in.2  ? 

115.  Compare  the  strength  of  a  pile  of  10  boards, 
each  14  ft.  long,  1  ft.  wide,  and  1  in.  thick,  when 
the  boards  are  piled  horizontally  and  when  they  are 
placed  close  together  on  edge. 

116.  Design  a  rectangular  wooden  cantilever  to     , 
project  4  ft.  from  a  wall  and  bear  a  load  of  500  Ib.     ( 
at  its  end,  the  factor  of  safety  being  8. 

117.  A  wooden  girder  supporting  the  bearing  par- 
titions in  a  dwelling  is  made  up  of  four  2-in.  by  10-in.  joists  set  on  edge  and  spiked 
together.   Find  the  size  of  a  steel  I-beam  of  equal  strength. 

118.  A  factory  floor  is  assumed  to  carry  a  load  of  200  Ib./f  t.2  and  is  supported 
by  steel  I-beams  of  16-f t.  span  and  spaced  4  ft.  apart  on  centers.   What  size  I-beam 
is  required  for  a  working  stress  of  16,000  lb./in.2  ? 


1 
1 

T 

ra 

G> 

•S.I.X 

i1 

}.  55 


58 


RESISTANCE  OF  MATERIALS 


119.  Find  the  required  size  of  a  square  wooden  beam  of  14-ft.  span  to  carry 
an  axial  tension  of  2  tons  and  a  uniform  load  of  100  Ib./ft. 

120.  A  floor  designed  to  carry  a  uniform  load  of  200  Ib./ft.2  is  supported  by 
10-in.  steel  I-beams  weighing  30  Ib./ft.    How  far  apart  may  they  be  placed  for 
a  span  of  16  ft.  and  a  working  stress  of  16,000  lb./in. 


=3—  T 

p 

D\ 

n—  ^     ' 

V 

\ 

^  * 

\ 

[l 

« 

-i  — 

h  A 

"-! 

f 

\    \ 

8g 

Shaft 

A   -- 

0 

) 

6=    !* 

Y 

J\  J 

£ 

-W 

± 

1  — 

\    i 

«-—  2Q--> 

3-3.-. 

r 

—  > 

r 

// 

J  

u^ 

PLAN 


END  ELEVATION 


FIG.  56 


121.  A  floor  is  supported  by  wooden  joists  2  in.  x  12  in.  in  section  and  16ft. 
span,  spaced  16  in.  apart  on  centers.    Find  the  safe  load  per  sq.  ft.  of  floor  area 
for  a  working  stress  of  800  lb./in.2 

122.  A  floor  is  required  to  support  a  uniform  load  of  150  Ib./ft.2  and  is  supported 
by  steel  I-beams,  18  ft.  span  and  spaced  5  ft.  apart  on  centers.    What  size  I-beam 

is   required   for  a   working 
stress  of  16,000  lb./in.2  ? 

123.  A      structural-steel 
built  beam  is  20  ft.  long  and 
has  the  cross  section  shown 
in  Fig.  53.    Compute  its  mo- 
ment of  resistance  and  find 
the  safe  uniform  load  it  can 
carry  per  linear  foot  for  a 
factor  of  safety  of  5. 

124.  The  cast-iron  bracket 
shown  in  Fig.  54  has  at  the 
dangerous  section  the  dimen- 
sions shown   in   the  figure. 
Find  the  maximum  concen- 
trated load  it  can  carry  with  a  factor  of  safety  of  15. 

125.  Find  the  proper  dimensions  for  a  wrought-iron  crank  of  dimensions  shown 
in  Fig.  55  for  a  crank  thrust  of  1500  Ib.  and  a  factor  of  safety  of  6. 

126.  A  wrought-iron  pipe  1  in.  in  external  diameter  and  TTg-  in.  thick  projects 
6  ft.  from  a  wall.   Find  the  maximum  load  it  can  support  at  the  outer  end. 


FIG.  57 


STRENGTH  OF  BEAMS 


59 


FIG.  58 


127.  The  yoke  of  a  hydraulic  press  used  for  forcing  gears  on  shafts  is  of  the 
form  and  dimensions  shown  in  Fig.  56.    The  yoke  is  horizontal,  with  groove  up, 

so  that  the  shaft  to  be  fitted  lies  in  the 
groove,  as  shown  in  plan  in  the  figure. 
The  ram  is  32  in.  in  diameter  and  under 
a  water  pressure  of  250  lb./in.2  Find 
the  dangerous  section  of  the  yoke  and 
the  maximum  stress  at  this  section. 

128.  A  10-in.  I-bar  weighing 40 Ib./ft. 
is  supported  on  two  trestles  15  ft.  apart. 
A  chain  block   carrying   a  1-ton   load 
hangs  at  the  center  of  the  beam.    Find 
the  factor  of  safety. 

129.  The  hydraulic  punch  shown  in 
Fig.  57  is  designed  to  punch  a  jj-in.  hole 
in  a  f-in  plate.    The  dimensions  of  the 

dangerous  section  AB  are  as  given  in  the  figure.  Find  the  maximum  stress  at 
this  section. 

130.  The  load  on  a  car  axle  is  8  tons,  equally  distributed  between  the  two 
wheels  (Fig.  58).    The  axle  is  of  cast  steel.    Find  its  diameter  for  a  factor  of 
safety  of  15. 

131.  The  floor  of  an  ordinary  dwelling  is 
assumed  to  carry  a  load  of  50  Ib./ft.2  and  is 
supported  by  wooden  joists  2  in.  by  10  in.  in 
section,  spaced  16  in.  apart  on  centers.   Find 
the  greatest  allowable  span  for  a  factor  of 
safety  of  10. 

132.  An  engine  shaft  of  machinery  steel 
rests  in  bearings  6  ft.  apart  between  centers 
and  carries  a  12-ton  flywheel  midway  between 
the  bearings.  Find  the  required  size  of  shaft. 

133.  A  cast-iron  flange  coupling  is  connected  with  ten  wrought-iron  bolts. 
Distance  from  axis  of  each  bolt  to  axis  of  shaft  is  6  in.    Total  torque  (twisting 
moment)  transmitted  is  12,000  ft.-lb.   If  the  flanges  are  accidentally  separated  2  in. 
and  the  bolts  are  a  drive  fit,  find  the  bending  stress  produced  in  each  bolt. 

134.  In  the  carriage  clamp  shown  in  Fig.  59  the  screw  is  of  wrought  iron,  |  in. 
diameter,  square  thread,  5  threads  per  inch,  and  the  casting  has  the  dimensions 

given  in  the  figure.  Find  what 
load  on  the  screw  will  cause  fail- 
ure by  shearing  the  threads,  and 
find  the  maximum  stress  in  the 
casting  under  this  load,  due  to 
combined  bending  and  tension. 
135.  In  the  joiner's  clamp 

shown  in  Fig.  60  the  bar  is  of  carbon  steel,  11  in.  x  Jin.,  tensile  strength 
70,000  lb./in.2,  and  the  screw  is  steel,  fin.  diameter,  square  threads,  5  threads 
to  the  inch.  Find  the  dimensions  of  the  cast-iron  handle  so  that  it  shall  be  light 
enough  to  act  as  the  breaking  piece, 


FIG.  59 


FIG.  60 


SECTION  V 

DEFLECTION  OF  CANTILEVER  AND  SIMPLE   BEAMS 

36.  General  deflection  formula.  By  a  simple  beam  is  meant  one 
which  is  simply  supported  at  the  ends.  The  only  external  forces 
acting  on  it  in  addition  to  the  loads  are,  then,  the  two  vertical  reac- 
tions at  the  supports.  A  cantilever  is  a  beam  which  overhangs,  or 
projects  outward  from  the  support,  the  loads  on  it  being  equili- 
brated by  the  moment  at  the  support  and  by  the  vertical  reaction 

at  this  point.  The  results  of 
applying  the  general  deflec- 
tion formula,  derived  below, 
to  these  two  classes  of  beams 
will  be  made  the  basis  of  the 
treatment  of  continuous  and 
restrained  beams  in  the  sec- 
tions which  follow. 

Taking  a  vertical  longi- 
tudinal section  of  a  beam, 
the  line  in  which  this  plane 
intersects  the  neutral-fiber 
surface  is  called  the  elastic 
curve.  Any  small  segment, 
A#,  of  the  elastic  curve  may  be  considered  as  a  circular  arc  with 
center  at  some  point  0  (Fig.  61).  This  point  0  is  therefore  called 
the  center  of  curvature  for  the  arc  Az.  The  radius  of  curvature  is 
not  constant,  but  changes  from  point  to  point  along  the  beam. 
Evidently  the  radius  of  curvature  is  least  where  the  beam  is  curved 
most  sharply. 

Any  two  adjacent  plane  sections,  AB  and  DH  (Fig.  61),  origi- 
nally parallel,  intersect  after  flexure  in  the  center  of  curvature  0. 
Let  KC  =  Az  denote  the  original  length  of  the  fibers,  and  draw 


DEFLECTION  OF  CANTILEVER  AND  SIMPLE  BEAMS      61 

through  C  a  line  EF  parallel  to  AB.  Then  DF  denotes  the  short- 
ening of  the  extreme  fiber  on  one  side,  and  EH  the  lengthening  of 
the  extreme  fiber  on  the  other.  Now,  since  the  triangles  KOC  and 
ECH  are  similar,  we  have  the  proportion 

EH  _  CH     e 

~KC~~OK~~r 

77*77" 

Moreover,  the  left  member, ,  is  the  change  in  length  of  the 

KG 

extreme  fiber  divided  by  its  original  length,  which  is  by  definition  the 

unit  deformation  s  of  this  fiber.  Also,  by  Hooke's  law,  —  =  E,  where 
jo  denotes  the  unit  nor- 
mal stress  on  this  fiber. 
Hence  the  above  pro- 
portion becomes 

p      e 

M     Me 
or,  since  p  =  —  =  — , 

as  shown  in  the  pre- 
ceding section, 

Me      e 


whence 
<35> 


F.o.62 


Now  let  AB  denote  any  segment  of  the  elastic  curve,  and  AAf,  BB' 
the  tangents  at  A  and  B  respectively  (Fig.  62).  If  AB  is  divided 
up  into  small  segments  A#,  and  A<£  denotes  the  angle  which  each 
subtends  at  the  center  of  curvature  0,  as  shown  in  Fig.  62,  then 

Az  =  7*A<£,  or  A</>  =  — ,  and,  inserting  in  this  the  value  of  r  obtained 
above,  it  becomes 


El 


Hence,  by  summation,  the  total  angular  deflection  </>  is 


62  RESISTANCE  OF  MATERIALS 

Now  for  any  small  arc,  A#,  the  deflection  Ac?  at  any  point  at  a  dis- 
tance #,  measured  from  the  tangent  to  the  arc  at  the  initial  point 
(Fig.  62),  is 

AcZ  = 


Hence  the  total  deflection  for  any  finite  portion  of  the  arc  AB, 
measured  from  one  end  A  to  the  tangent  at  the  other  end  B,  is 


But  M&x  denotes  the  area  of  a  small  vertical  strip  of  the  moment 
diagram  of  altitude  M  and  base  Arc,  and  V(JfA#)#  is  the  sum  of 
the  static  moments  of  all  these  elements  of  area  with  respect  to  the 
point  A.  From  the  results  of  Section  II,  however,  this  is  equal  to 
the  area  of  the  moment  diagram  between  A  and  B  multiplied  by 
the  distance  of  its  centroid  from  A.  That  is,  if  Aab  denotes  the  area 
of  the  moment  diagram  between  the  points  A  and  B,  and  XQ  is  the 
distance  of  the  centroid  of  Aab  from  A,  then 


Therefore  d  =  — -_  Aab  .  x, 

Jbl 

or,  in  general, 

(36)    d  =  -  -  (static  moment  of  the  moment  diagram). 
El 

The  angular  deflection  $  between  any  two  points  A  and  B,  that 
is,  the  angle  between  the  tangents  to  the  elastic  curve  at  these  two 
points,  is  given  by 


Therefore,  since  ^  Mt±x  denotes  the  area  A^  of  the  moment  diagram 
between  the  two  points  in  question,  and  since  for  the  small  deflec- 
tions which  actually  occur  in  practice  we  may  assume  <f>  =  tan  cf> 
without  introducing  any  appreciable  error, 


(37) 

El 


DEFLECTION  OF  CANTILEVER  AND  SIMPLE  BEAMS      63 


37.  Cantilever  bearing  concentrated  load.    For  a  cantilever  bear- 
ing a  concentrated   load   at  the  end,  the  moment  diagram  is  a 

triangle,  as  shown  in 
Fig.  63.  The  area 
of  the  moment  dia- 
gram is  therefore 

PI2 

A  =  ——,  and  the  dis- 
tance of  its  centroid 
from  the  free  end 

2 

is  XQ  =  -  L  Therefore 
o 

the  deflection  at  the 
free  end  is 


FIG.  63 


(38) 


d  = 


E  I 


PI* 
- 
3  El 


The  deflection  d  may  also  be  expressed  in  terms  of  the  stress 

on  the  extreme  fiber.    Thus,  since  %-—=M=Pl,   by  substituting 
this    value    of  PI   in 
the  expression  for  c?,  it 
becomes 


(39) 


3Ee 


Also,  the  angular 
deflection  at  the  load 
is  found  to  be 


(40)   tan^  =  _ 


M= 


FIG.  64 


A 
El 

_     PI* 
~  2  El 

If  the  load  is  at  a  distance  a  from  the  fixed  end  and  b  from  the 
free  end  (Fig.  64),  then  the  deflection  at  the  load,  as  shown  above,  is 

Pas 


(41) 


d  = 


64 


RESISTANCE  OF  MATERIALS 


and  similarly,  from  equation  (40),  the  angular  deflection  at  the  load  is 

fa2 
~  2  El' 


(42) 


Consequently,  the  additional  deflection  d'  from  the  load  to  the  free 
end  of  the  cantilever  is 


(43) 


df  =  b  tan  <j>  = 


2  El 


The  total  deflection  D  at  the  free  end  is  therefore 


It  is  often  convenient  to  let  b  =  M,  where  k  denotes  a  proper 
fraction.  Then  in  the  present  case  a=  I—  b  =  l(~L  —  &),  and  the 
expression  for  the  deflection  at  the  end  becomes 

p/3 


(45) 


For  instance,  if  the  load 
is  at  the  middle  of  the 
cantilever,  then  k  =  i, 
and  the  deflection  at 
the  free  end  becomes 


38.  Cantilever  bear- 
ing uniform  load.  For 
a  uniformly  loaded 
cantilever  the  moment 
diagram  is  a  parabola 
and  the  moment  at 


the  support  is  M=wl>-  —  ^-  (Fig.  65).    Also,  from  article  17, 

-f         72  73 

the  area  of  the  moment  diagram  is  A  =  -  —  -  •  I  =  —  ,  and  the  dis- 

o   2        Q    " 

tance  of  its  centroid  from  the  free  end  is  XQ  =  -I.    Therefore  the 
deflection  at  the  free  end  is 


(47) 


DEFLECTION  OF  CANTILEVER  AND  SIMPLE  BEAMS      65 

From  the  relation  *—  =  M  =  —  the  expression  for  the  deflection 
in  terms  of  the  maximum  fiber  stress  p  is  found  to  be 


C")  d  =  fWe 

Also,  the  total  angular  deflection  at  the  end  of  the  beam  is 

(49) 


A          wl3 

tan  ©  = = 

El       GEI 


39.  Cantilever  under  constant  moment.  If  a  cantilever  is  sub- 
jected to  a  couple,  that  is,  a  pair  of  equal  and  opposite  parallel 
forces,  as  shown  in  Fig.  66,  the  moment  is  constant  for  the  entire 
length  of  the  beam.  The  moment  diagram  is  therefore  a  rectangle 
of  area  Ml,  and  the  de- 
flection at  the  free  end  is 


(50)       d  =  — 


The  angular  deflection 
at  the  free  end  in  this 


case  s 


Ml 


(51) 


40.  Simple  beam  bearing  concentrated  load.  To  apply  the  deflec- 
tion formula  to  a  simple  beam,  the  deflection  must  be  measured  at 
one  end  A  from  a  tangent  at  the  middle  C.  For  a  concentrated 
load  P  at  the  middle  (Fig.  67),  the  area  of  the  moment  diagram 

7372  2         I  I 

from  A  to  C  is  A  =  -^-,  and  x.  =  -.-  =  -.    Hence,  in  this  case 

Ib  O       A          O 


(52) 


Fl 


66  RESISTANCE  OF  MATERIALS 

Also,  the  total  angular  deflection  for  half  the  beam  is  found  to  be 

(53) 


.       A        pi2 

tan  9  = = 

EI      !<>/•;/ 


The  deflection  may  also  be  obtained  by  considering  the  moment 
diagram  as   representing  the  load  on  the  beam,  and  then  taking 

moments  about  the 
point  at  which  the 
deflection  is  meas- 
ured, say  the  center 
C  (Fig.  67).  Since 
the  total  area  of  the 
moment  diagram  is 
1  n  j_PP 

is  regarded  as  the 
load  on  the  beam 
each  reaction  will  be 

PI'2 

—  •     Then,    taking 


Fi«.  07 
moments  about  the  center,  the  result  is 

~ 


KJ 


i    pp   i 

2  ~  Hi"  '  ( 


PI* 


48  El 


' 


From  the  relation  ^—  =  M=  — ',  the  deflection  at  the  center  may 
e  4 

be  expressed  in  terms-  of  the 
maximum  liber  stress  p.  Thus, 

PI  .  .  pi 

replacing  -—  by  its  equal  - 

in  the  expression  for  <7,  the 
result  is 


(55) 


d  = 


16 


16 


FIG.  68 


If  the  concentrated  load  P  is  not  at  the  center  but  divides  the 
span  /  into  two  unequal  segments  a  and  £>,  the  reactions  are  — ,  —-, 


DEFLECTION  OF  CANTILEVER  AND  SIMPLE  BEAMS     67 


and  the  moment  at  the  load  is  — — .    Also,  the  area  of  the  moment 

diagram  between  the  load  and  one  end  is  -  •  - — ,  and  the  distance 

.2 

of  the  centroid  of  this  segment  from  the  end  is  :r  =  ^  a.    Hence  the 

o 

deflections  of  the  ends  from  the  tangent  at  the  point  of  application 
C  of  the  load  are  j,^  ,,  .7!5 


ZEIl  ZKli 

and  the  deflection  of  C  below  the  level  of  the  supports  is 

I'uW        pah 


(56) 


XEIl 


where  p  denotes  the  maximum  fiber  stress. 

41.   Simple  beam  bearing  uniform  load.    For  a  simple  beam  uni- 
formly loaded  the  moment  diagram  is  a  parabola,  the  maximum 

wl* 
ordinate   being  -  — • 

8 

From  article  17,  the 
area  of  this  parabola  is 
2    wl     -.      wl 

A=»'T'l=w 

To  apply  the  general 
formula  for  deflec- 
tion, consider  d  as 
measured  from  one 
end  A  to  the  tan- 
gent at  the  center 
C  (Fig.  ()(.)).  Then, 
since  the  area  of  one 
half  the  moment  diagram  is  -,  and  the  distance;  of  the  centroid  of 

f>     /       f>  / 
this  half  from  a  vertical  through  A  is  XQ  -   -  •      —  -  - . »  tlie  deflection  is 

8     ii      KJ 


(57) 


— 

El 


El 


To  express  the  deflection  in  terms  of  the  maximum  fiber  stress  />, 

make  use  of  the  relation  ^—  =  M=  ^—.    Then,  replacing  -— ,  in  the 

e  o  o 


68  RESISTANCE  OF  MATERIALS 

expression  for  d,  by  its  equal  ±—  ,  it  becomes 

«> 

The  total  angular  deflection  for  half  the  beam  in  this  case  is 

A  wl* 

(59}  tan  9  =  -  =  -  . 

El       24  E  I 

The  deflection  may  also  be  obtained  by  regarding  the  moment 
diagram  as  representing  the  load  on  the  beam.    Since  the  total  area 

wls  wl8 

of  the  moment  diagram  is  —  ,  each  reaction  will  then  be  -—  ,  and 

\2t  2<± 

therefore,  taking  moments  about  the  center  to  find  the  deflection  at 
this  point,  the  result  is,  as  before, 

,__!_  /wP    L_wP     3    \        5  wl* 
~  '  ' 


2      24     16          884JBT 


APPLICATIONS 

136.  A  15-in.  .I-beam  weighing  60  Ib./ft.  carries  a  25-ton  load  at  the  center  of 
a  12-ft.  span.   Find  the  maximum  deflection. 

137.  In  building  construction  the  maximum  allowable  deflection  for  plastered 
ceilings  is  ^^  of  the  span.   A  floor  is  supported  on  2  in.  x  10  in.  wooden  joists 
of  14-ft.  span  and  spaced  16  in.  apart  on  centers.    Find  the  maximum  load  per 
square  foot  of  floor  surface,  in  order  that  the  deflection  may  not  exceed  the 
amount  specified. 

138.  Determine  the  proper  spacing,  center  to  center,  for  12-in.  steel  I-beams 
weighing  351b./ft.,  for  a  span  of  20ft.  and  a  uniform  floor  load  of  100  Ib./ft.2,  in 
order  that  the  deflection  shall  not  exceed  ^^  of  the  span. 

139.  A  structural  steel  shaft  8  in.  in  diameter  and  5  ft.  long  between  centers  of 
bearings  carries  a  25-ton  flywheel  midway  between  the  bearings.   Find  the  maxi- 
mum deflection  of  the  shaft,  considering  it  as  a  simple  beam. 

140.  A  wrought-iron  bar  2  in.  square  is  bent  to  a  right  angle  4ft.  from  one 
end.  The  other  end  is  then  embedded  in  a  concrete  block  so  that  it  stands  upright 
with  the  4ft.  length  horizontal.    If  the  upright  projects  12  ft.  above  the  concrete, 
and  a  load  of  300  Ib.  is  hung  at  the  end  of  the  horizontal  arm,  find  the  deflection 
at  the  end  of  this  arm. 

141.  A  wooden  cantilever  2  in.  x  10  in.  in  section,  with  the  longer  side  vertical, 
projects  10  ft.  from  the  face  of  a  wall  and  carries  a  concentrated  load  of  600  Ib.  at 
a  point  6  ft.  from  the  wall.   Find  the  deflection  at  the  free  end  of  the  beam. 

142.  A  10-in.  steel  I-beam  weighing  40  Ib./ft.  spans  an  opening  16  ft.  wide  and 
supports  a  total  load  of  40  tons.    Find  how  much  greater  the  maximum  deflection 
of  the  beam  is  when  this  load  is  concentrated  at  its  center  than  when  it  is  distri- 
buted uniformly  over  the  beam. 


DEFLECTION  OF  CANTILEVER  AND  SIMPLE  BEAMS      69 

143.  A  built  beam  is  composed  of  two  10-in.  steel  channels,  40  lb./ft.,  placed  on 
edge  and  connected  with  latticing.   The  span  is  20  ft.   Find  what  uniform  load  per 
linear  foot  the  beam  can  carry  under  the  condition  that  the  maximum  deflection 
shall  not  exceed  ^  in. 

144.  A  15-in.  steel  I-beam,  42  lb./ft.,  spans  an  18-ft.  opening.    Find  the  maxi- 
mum deflection  for  a  maximum  fiber  stress  of  16,000  lb./in.2 

145.  The  total  load  on  a  car  axle  is  10  tons,  equally  distributed  between  the  two 
wheels.  Distance  between  centers  of  wheels  is  56  in.,  and  between  centers  of  bear- 
ings is  68  in.  Find  the  maximum  deflection  of  the  axle  measured  from  a  horizontal 
line  joining  the  centers  of  bearings. 

146.  A  10-in.  steel  I-beam  weighing  30  lb./ft.  rests  on  two  supports  16  ft.  apart 
and  carries  a  uniform  load  of  200  lb./ft.  in  addition  to  its  own  weight.    A  third 
support  just  touches  the  beam  at  the  center.    How  much  must  this  central  support 
be  raised  so  that  it  shall  carry  all  the  weight,  and  the  beam  just  touch  the  two  end 
supports  ? 

147.  A  cast-iron  pipe  20  in.  internal  diameter  and  1  in.  thick  rests  on  supports 
30  ft.  apart.   Find  the  maximum  deflection  when  the  pipe  is  full  of  water. 

148.  A  beam  of  uniform  section,  carrying  a  concentrated  load  at  the  center, 
has  a  maximum  deflection  equal  to  1  per  cent  of  the  span.    Find  the  slope  of  the 
beam  at  its  ends. 

149.  Three  beams  of  the  same  material  are  laid  side  by  side  across  an  opening 
of  12-ft.  span,  and  a  load  of  1000  Ib.  rests  across  them  at  the  center  of  the  span  so 
that  they  must  all  bend  together.    The  beams  are  each  2  in.  wide,  but  two  of  them 
are  6  in.  deep,  while  the  third  is  12  in.  deep.    How  much  of  the  weight  is  carried 
by  each  beam  ? 

150.  A  steel  bar  2  in.  square  rests  on  knife  edges  5  ft.  apart,  and  its  maximum 
deflection  under  a  central  load  of  1000  Ib.  is  found  to  be  .1125  in.    Calculate  from 
this  experiment  the  modulus  of  elasticity  of  the  bar. 


SECTION  VI 

CONTINUOUS  BEAMS 

42.  Theorem  of  three  moments  for  uniform  loads.  A  continuous 
beam,  or  girder,  is  one  which  is  supported  at  several  points  of  its 
length.  The  reactions  and  moments  in  this  case  are  statically 
indeterminate ;  that  is  to  say,  the  ordinary  static  conditions 
of  equilibrium,  ^f=  0,  ^Tj/  =  0,  are  insufficient  to  determine 
them.  To  solve  the  problem  it  is  necessary  also  to  take  into 
account  the  deflections  of  the  beam. 


FIG.  70 

The  simplest  method  of  finding  the  reactions  and  moments 
at  the  supports  for  a  continuous  beam  is  by  applying  what  is 
known  as  the  theorem  of  three  moments.  This  theorem  establishes 
a  relation  between  the  moments  at  three  consecutive  supports 
of  a  continuous  beam  and  the  loads  on  the  two  included  spans, 
and  was  first  published  by  Clapeyron  in  1857.  The  following 
proof  of  the  theorem,  however,  is  very  much  simpler  than  any 
previously  given, 

70 


CONTINUOUS  BEAMS 


71 


For  a  continuous  beam  bearing  a  uniform  load  let  A,  B,  C  denote 
any  three  consecutive  points  of  support,  assumed  to  be  in  the  same 
line,  and  let  M^  M^  M3 ;  JK1?  R^  E9  denote  the  moments  and  reac- 
tions at  these  three  points  respectively.  Also  let  l^  Z2  denote  the 
lengths  of  the  two  spans  considered,  w^  w2  the  unit  loads  on  them, 
and  /Sff,  /Sf  the  shears  on  the  right  and  left  of  E^  respectively 
(Fig.  70),  with  a  similar  notation  for  the  other  points  of  support. 

Now  consider  a  portion  of  the  beam  cut  off  by  planes  just  inside 
the  supports  at  A 
and  (7,  as  shown  in 
Fig.  71.  Then,  con- 
sidering the  end  B 
as  fixed,  the  deflec- 
tion at  A  from  the 

tangent  at  B  consists  of  three  parts :  that  due  to  the  moment 
to  the  shear  S?  considered  as  a  load,  and  to  the  uniform  load 
Calling  these  deflections  d^  c?2,  t?3,  respectively,  we  have 


(Eq.  (50),  Art.  39) 
(Eq.  (38),  Art.  37) 


CifijS 

\  li 


SIS  I 


(Eq.  (47),  Art.  38) 


Hence  the   total  deflection  DA  of  the  point  A  measured  from  a 
tangent  at  the  point  B  is 


(61) 


DA  =  - 


To  eliminate  the  shear   £f,  form  a  moment   equation  by  taking 
moments  about  the  point  B.    Then 


whence 
(62) 


2   '• 


72  RESISTANCE  OF  MATERIALS 

and,  substituting  this  value  of  $f  in  the  above  expression  for  DA,  it 
reduces  to 


Similarly,  by  considering  the  span  BC  and  calculating  the  deflec- 
tion Dc  of  the  point  C  measured  from  the  same  tangent  at  B,  we 
obtain  the  equation 


D  - 


Also,  forming  a  moment  equation  with  C  as  center  of  moments, 

we  have  ™ 

j|fa  =  j|f8-$7a +  _|i., 

and,  eliminating  /Sf  between  these  relations,  the  result  is 
(65)       •  Dc=-^jL 


Now,  since  these  deflections  lie  on  opposite  sides  of  the  tangent 
at  B,  we  have,  from  similar  triangles, 


Therefore,  substituting  the  expressions  for  DA  and  Dc  in  this  rela- 
tion, combining  terms,  and  transposing,  we  obtain  the  relation 


(66)  M&  +  2  M2  ft  +  Q  +  M^  =  . 

In  this  relation,  M^  M2,  and  MB  are  stress  couples  acting  on  the  beam. 
The  external  moments  at  the  supports  are  equal  in  amount  but 
opposite  in  sign  to  the  stress  couples,  or  internal  moments.  There- 
fore, calling  Jfj,  Jf2,  Ms  the  external  moments  at  the  supports,  the  sign 
of  the  expression  is  changed  ;  that  is 


(67) 


This  is  the  required  theorem  of  three  moments  for  uniform  loads. 

43.  Theorem  of  three  moments  for  concentrated  loads.    Consider 
a  continuous  beam  bearing  a  single  concentrated  load  in  each  span. 


CONTINUOUS  BEAMS 


73 


The  distance  of  the  load  in  any  span  from  the  adjoining  support  on 
the  left  will  be  denoted  by  M,  where  I  is  the  length  of  the  span  and 
k  is  a  proper  fraction ;  that  is,  kl  is  some  fractional  part  of  the  span 
(Fig.  72).  Thus,  if  the  load  is  at  the  middle  of  the  span,  k  =  i  ;  if 
it  is  at  the  quarter  point,  k  =  ^,  etc. 

Now  consider  a  portion  of  the  beam  extending  over  three  con- 
secutive supports  A,  J5,  and  (7,  and  let  M^  M^  Jf3  denote  the 
moments,  and  R^  R^  R^  the  reactions,  at  these  supports.  Then,  to 
obtain  the  theorem  of  three  moments,  calculate  the  deflections  of 
A  and  C  measured  from  the  tangent  to  the  elastic  curve  at  B.  To 
calculate  the  deflection  of  A,  suppose  the  beam  to  be  cut  by  a  plane 
just  inside  the  sup- 
port at  A,  and  call 
the  shear  on  the 
section  S*.  Then, 
considering  the  end 
B  as  fixed,  calculate 
the  deflection  of  A 
by  treating  the  part  AB  as  a  cantilever  subjected  to  the  moment  M^ 
the  shear  $f  regarded  as  a  load,  and  the  concentrated  load  J^.  Call- 
ing these  three  partial  deflections  d^  d^  d^  respectively,  we  have 


(Eq.  (50),  Art.  39) 


(Eq.  (38),  Art.  37) 


, 


3  El      2  El 


(Eq.  (44),  Art.  37) 


In  the  present  notation  the  quantities  a  and  b  in  the  expression  for 

dz  are 

a  =  distance  from  fixed  end  =  \  —  k^, 

b  =  distance  from  free  end    =  k^. 
Substituting  these  values  of  a  and  £,  the  equation  for  d3  becomes 

T>73 

/"OQ"\  fl    1  1      /^O  O   7n      i     i"3"\ 

v°°y  8  £?77irv  i"iiy* 


74  RESISTANCE  OF  MATERIALS 

Therefore,  by  addition,  the  total  deflection  of  the  end  A  with  respect 
to  the  tangent  at  B  is 


Now,  forming  a  moment  equation  for  the  portion  AB,  taking  center 
of  moments  at  B,  we  have 


whence 


S*  =     (Jf,  -  Jf2)  +  ^(1  -  &,), 


and  eliminating  /Sf  between  this  equation  and  the  expression  for 


DA,  the  result  is 


Similarly,  to  find  the  deflection  at  (7,  measured  from  the  tangent 
to  the  elastic  curve  at  J5,  treat  the  portion  BC  as  a  cantilever  fixed 
at  B  and  subjected  to  the  moment  M^  the  shear  >Sf  considered  as 
a  load,  and  the  concentrated  load  ^.  Then,  calling  these  partial 


deflections  dl9  d2,  d^  we  have 


(Eq.  (50),  Art.  39) 
(Eq.  (38),  Art.  37) 


d>=-+  (Eq-  (44X  Art.  37) 


l 

or,  since  in  the  present  case  a  =  kj,2,  b  =  Z2(l  —  &2),  the  expression 
for  dg  becomes 

(71) 


Therefore  the  total  deflection  Dc  from  the  tangent  at  B  is 


CONTINUOUS  BEAMS  75 

Now,  forming  a  moment  equation  for  the  portion  BC,  taking  center 
of  moments  at  B,  we  have 


whence 

(73)  Sf  = 

2 

and,  eliminating  /Sj  between  this  equation  and  the  expression  for 
Z>c,  the  result  is 


(74)         Dc  =  -  =-  -  p     4.         _  (2  &       3  &22  +  &23). 
6  J£T      ZEISEI^ 

Since  the  deflections  at  ^4  and  C  lie  on  opposite  sides  of  the  tan- 
gent at  B,  we  have,  from  similar  triangles, 


Substituting  in  this  relation  the  values  of  DA  and  Dc  just  found, 
combining  like  terms,  and  transposing,  we  obtain  the  relation 


(75) 

2-  8 


In  this  relation  Jlf1?  Mz,  M3  are  the  stress  couples  acting  on  the 
beam.  The  external  moments  at  the  supports  are  equal  in  amount 
but  opposite  in  sign  to  the  stress  couples.  Therefore,  calling  M^  M^ 
Mz  the  external  moments  at  the  supports,  the  sign  of  the  expression 
is  changed  ;  that  is, 

(76)   M&  +  2  M&i  +  12)  +  MZ12  =  -  PJ*fa  -  kl) 


which  is  the  required  theorem  of  three  moments  for  a  single  con- 
centrated load  in  each  span. 

For  a  single  concentrated  load  at  the  center  of  each  span,  each 
k  =  i.    In  this  case  the  theorem  becomes 


(77)        . 

o 

If  there  are  a  number  of  concentrated  loads  in  each  span,  an 
equation  like  (76)  can  be  written  for  each  load  separately.    By 


76  RESISTANCE  OF  MATERIALS 

adding  these  equations  the  general  theorem  of  three  moments  for 
any  number  of  concentrated  loads  is  found  to  be 

(78)    MJt  +  2  M2(l,  +  12)  +  M312  =  -  ]?lVia(*i  -  *') 

-]£lV.'(2fc,-  3  *;  +  *»). 

44.  Effect  of  unequal  settlement  of  supports.  In  deriving  the 
theorem  of  three  moments  the  supports  were  assumed  to  be  at  a  fixed 
elevation  in  the  same  line.  If  their  relative  elevation  changes,  owing 
to  unequal  settlement  of  the  supports  or  to  other  causes,  the  effect 
in  general  is  to  increase  the  stress  in  the  member.  To  take  account 
of  this  effect  in  applying  the  theorem,  suppose  that  the  supports 
were  originally  in  line,  and  denote  the  settlement  of  three  consecu- 
tive supports  A,  B,  C  from  their  original  level  by  h^  h2,  h3,  respec- 
tively. Then  the  difference  in  elevation  between  A  and  B  is  lil  —  h^ 
and  between  B  and  C  is  h^  —  li^.  Thus,  if  A  settles  more  than  B, 
\  ~  \  *s  Positive  and  the  deflection  at  A  is  increased  by  this 
amount.  If  A  settles  less  than  B,  Ji^  —  h2  is  negative  and  the 
deflection  at  A  is  decreased  by  this  amount,  etc.  In  general,  then, 
equations  (70)  and  (74)  for  the  deflections  at  A  and  C  become 


Substituting  these  values  of  DA  and  Dc  in  the  relation 


the  result,  after  combining  terms  and  changing  the  signs  of 
M,  M,  is 


=  -  2)  pJi  (*i  -  *')  -  S  r^ 


CONTINUOUS  BEAMS  77 

This  relation  is  therefore  the  most  general  form  of  the  theorem  of 
three  moments  for  any  number  of  concentrated  loads,  including  the 
effect  of  unequal  settlement  of  the  supports,  or  other  change  in 
their  relative  elevation. 

APPLICATIONS 

151.  A  continuous  beam  of  four  equal  spans  is  uniformly  loaded.   Find  the 
sending  moments  at  the  supports. 

Solution.    The  system  of  simultaneous  equations  to  be  solved  in  this  case  is 

M1  =  M5  =  0, 


7/,/2 

~ 


the  solution  of  which  gives 


152.  A  continuous  beam  of  n  equal  spans  carries  a  uniform  load  of  the  same 
amount  in  each  span.  Check  the  moments  at  the  supports  given  in  the  following 
table  for  values  of  n  from  2  to  7.  The  tabular  values  here  given  are  the  numerical 
coefficients  of  —  wl2. 


MOMENTS  AT  SUPPORTS  FOR  EQUAL  SPANS  AND  UNIFORM  LOAD 


0 

0 

0 

0 

8 

0 

0 

0 

0 

t 

1 

ft 

t 

2 

l 

To 

t 

! 

1 
JL 

28 

t 

2 

2 

28 

t 

3 

3 

28 

t 

t 

A 

t 

2 

i 

t 

3 

I 

4 

38" 

i 

1 

0 

1 

* 

I 

2 

t 

3 

9 

ToT 

f 

4 

t 

5 

A 

t 

0 

0                   ft 

1 

2 

ft 

1 

3 

12 
T42 

1 

4 

I 

5 

ft 

1 

:,  i 

142                          0 

78  RESISTANCE  OF  MATERIALS 

SHEARS  AT  SUPPORTS  FOR  EQUAL  SPANS  AND  UNIFORM  LOAD 

1 


2 
1 

^k 

2 
2 

.oJjL 

8 
I 

5J5                    3)0 
8                         8 
2                     3 

.o|*               e|s 

10                       10 

1                   2 

±il           _iLl 

10                        10 

3                   4 

p|ll                 17J15 
28                        28 
1                     2 

13[l3                  is|l7                  ll|o 
28                       28                       28 
345 

of  15                   23J20                   18J19 
38                        38                         38 

1                     2                    3 

19J18                   20J23                   is[o 
38                         38                        38 

456 

o|41                   63J55                  .49  1  51 
104                       104                       104 

123 

53|53                   51  J49                   55|  63                   41JO 
104                       104                      104                       104 

4567 

o|56                  86|75                  67  170                  72  J71 
142                       142                      142                       142 

71J72                  70|67                  7o|86                  56J  0 
142                       142                       142                       142 

153.    Calculate  the  reactions  of  the  supports  in  problem  151. 

Solution.  The  reaction  at  any  support  may  be  found  by  finding  the  shears  close 
to  the  support  on  each  side.  The  sum  of  these  two  shears  is  then  equal  to  the  re- 
action. Thus,  in  the  present  case,  to  find  any  given  reaction,  say  #2,  consider  the 
portion  of  the  beam  between  Rl  and  R2,  as  shown  in  Fig.  70,  and  form  the  moment 
equation  for  this  segment.  Then 


and  therefore,  since  Ml  =  0  and  M2  =  ^3g  wZ2,  Sf  =  ^|  wl.    Similarly,  the  moment 
equation  for  the  segment  of  the  beam  between  R2  and  R3  is 


whence,  by  substituting~3f3  =  ^  wP  and  M2  =  ^  wZ2,  we  have 


Consequently,  E2  =  8}  +  Sf  =  wl  =  —  wl. 

28  28 

This  method  applies  when  it  is  required  to  find  one  reaction  only,  independently 
of  the  others.  If  all  the  reactions  are  required,  it  is  simpler  to  calculate  them  in 
succession,  starting  at  one  end,  without  reference  to  the  shears.  For  instance,  to 
find  JR1,  take  a  section  through  R2  and  consider  the  loads  on  the  left  of  the  section. 
Then  the  moment  equation  for  this  portion  is 


whence 


CONTINUOUS  BEAMS  79 

To  find  JR2  ,  take  a  section  through  R3  and  consider  all  the  loads  on  the  left  of  the 
section.   Then  the  moment  equation  is 


and  inserting  the  value  of  Bl  just  obtained,  it  is  found  that 


By  this  method  each  reaction  may  be  obtained  in  terms  of  those  already  found, 
without  calculating  the  shears. 

154.  In  problem  152  determine  the  shears  and  reactions  at  the  supports  and 
check  the  results  with  the  values  tabulated  on  page  78.    The  tabular  values  are 
the  numerical  coefficients  of  wL 

155.  An  18-in.  steel  I-beam,  601b./ft.,  is  continuous  over  four  supports,  the 
lengths  of  the  three  spans,  beginning  at  the  left,  being  25ft.,  40ft.,  and  35ft., 
respectively.    What  uniform  load  per  foot  run  would  produce  a  maximum  fiber 
stress  in  the  beam  of  16,000  lb./in.2  ? 


SECTION  VII 


RESTRAINED,  OR  BUILT-IN,  BEAMS 

45.  Uniformly  loaded  beam  fixed  at  both  ends.  By  a  restrained, 
or  built-in,  beam  is  meant  one  which  is  fixed  in  direction  at  certain 
points  of  its  length,  usually  the  ends  —  as,  for  example,  beams  built 
into  a  wall  or  forming  a  part  of  monolithic  concrete  construction. 
The  simplest  form  of  restrained  beam  is  a  cantilever,  which  can  be 
treated  by  ordinary  methods,  as  explained  in  articles  37,  38,  and  39. 

Consider  first  a  uni- 
formly loaded  beam  fixed 
at  both  ends,  as  shown  in 
Fig.  73.  Let  B,  F  denote 
the  points  of  inflection 
of  the  elastic  curve ;  that 
is,  the  points  at  which 
the  bending  moment  is 
zero.  Then  the  central 
portion  BF  may  be  con- 
sidered as  a  simple  beam 
of  length  2  x  bearing  a 
total  uniform  load  of 
FIG.  73  amount  2  wx,  and  each 

of  the  ends,  AB  and  FE, 

as  a  cantilever  uniformly  loaded  and  carrying  a  concentrated  load 
wx  at  the  end,  equal  to  one  of  the  reactions  for  the  portion  BF. 

If,  then,  d  denotes  the  deflection  of  the  point  F  with  respect  to 
A  or  E,  assumed  to  be  at  the  same  level,  the  value  of  d,  computed 
from  the  segment  AF,  is,  from  (38)  and  (47),  articles  37  and  38, 


(80) 


d  = 


1 


80 


BESTBAINED,  OB  BUILT-IN,  BEAMS  81 

and,  computed  from  the  segment  FE,  is 

w/i__  v 

(81)  d=     \?/    +  - 


Equating  these  two  values  of  the  deflection  d  and  solving  for  #,  the 
result  is  7 


2V3 

The  length  of  the  central  portion  BF  is  therefore  2  x  =  — - ,  and 
the  maximum  moment,  which  occurs  at  the  center  (7,  is 

iv I2 
(82)  Mc  =  —  • 

Similarly,  the  negative  moment  at  the  support  A  or  A  is 


2      /          a 

and,  since  a;  = 7=  >  this  reduces  to 

2V3 

(83)  MA=-^L. 

The  maximum  deflection  for  the  central  portion  BF,  considered 
as  a  simple  beam,  is,  from  (57),  article  41, 


V3/  9 

djrF"     384  AY    "384  A 


and  for  one  end,  say  AB,  considered  as  a  cantilever,  is,  from  (38) 
and  (47),  articles  37  and  38, 

I         *    \*        4 

-wl* 


.        -     -,        -     -   2      2V3/       _9 

-""  8  AY  3  AY  384  AY* 

Therefore,  since  the  total  deflection  of  the  center  C  below  the  sup- 
ports at  A  or  A  is  the  sum  of  these  two,  we  have 


82 


RESISTANCE  OF  MATERIALS 


46.  Beam  fixed  at  both  ends  and  bearing  concentrated  load  at 
center.   Following  the  method  of  the  preceding  article,  let  B  and  D 

denote  the  points  of  in- 
flection of  the  elastic 
curve,  or  positions  of 
zero  moment  (Fig.  74). 
Then  the  equilibrium 
would  not  be  disturbed 
if  the  beam  was  hinged  or 
jointed  at  B  and  D,  and 
it  may  therefore  be  con- 
sidered as  a  simple  beam 
of  length  BD  suspended 
from  the  ends  of  two  can- 
tilevers AB  and  DE. 

Now  consider  the  seg- 
ment AD  and   compute 
from  (44),  article  37,   the 


the  deflection    of    D  below  A.    Then, 
deflection  at  D  due  to  the  load  P  is 


(87) 


Pa2  /a 


El  V3      2 


where  in  the  present  case  a  =  —  and  b  =  x,  and  consequently 


(88) 


dp  = 


But  from  (38),  article  37,  the  load  — ,  acting  upward  at  D,  produces 
a  deflection  upward  of  amount 


P 


(89) 


Consequently  the  total  deflection  of  D  below  A  is 


RESTRAINED,  OR  BUILT-IN,  BEAMS  83 

Similarly,  for  the  portion  DE,  the  deflection  of  D  below  E  is 

P 

(91)  dDE  =  —  ^ 

Equating  these  two  values  of  the  deflection  and  solving  for  a;,  the 
result  is 

and  consquently  the  length  of  the  central  portion  BD  is 

•.-j- 

Therefore  the  moment  at  the  center  C  and  also  at  each  end  is  of 

numerically  the  same  amount,  namely,  — ^-£ »  or 

(92)  3f  =  f. 

The  maximum  deflection  for  the  central  portion  BD  is  the  same 
as  for  a  simple  beam  of  span  - ,  namely, 


(93) 


and  for  either  end  AB  or  D^  is  the  same  as  for  a  cantilever  of 


I  P 

length  -  carrying  a  load  —  at  the  end,  namely, 


2  \4 
(94)  dAB  =  - 


3  7?/       384  El 

Therefore  the  total  deflection  of  the  center  C  below  the  level  of  the 
supports  at  A  and  D  is  the  sum  of  these  two,  that  is, 

(95) 


84 


RESISTANCE  OF  MATERIALS 


47.  Single  eccentric  load.  For  a  beam  fixed  at  both  ends  and 
bearing  a  single  concentrated  eccentric  load  the  simplest  method 
of  computing  the  unknown  reactions  and  moments  at  the  sup- 
ports is  as  follows: 

Consider  the  beam  as 
fixed  at  one  end  E  only 
(Fig.  75)  and  carrying,  in 
addition  to  the  concen- 
trated load  P,  the  shear  Rl 
at  the  left  support  and  the 
restraining  moment  Ml  at 
this  point.  Then,  from  (50) 
and  (51),  article  39,  the  de- 
flection from  the  tangent  at 
E  due  to  the  moment  MI  is 

MJ, 


From  (38)  and  (40),  article  37,  that  due  to  the  shear  R1  is 
,        Rf  Rf  . 


and  from  (42)  and  (44),  article  37,  that  due  to  the  load  P  is 

,  Pll       PIJ22 

d  =-— ^--_Ll,  tancf>=- 


Since  the  total  vertical  deflection  of  the  point  A  with  respect  to 
the  point  E  is  zero,  and  since  the  total  angular  deflection  is  also 
zero,  these  two  conditions  furnish  the  equations 


(96) 


Mf       Rf 
2  El  +  3  El 


PIJ? 
2  El 


R,l2        Pll 


2  El      2  El 


=  0. 


From  the  second  equation, 


PI* 


RESTRAINED,  OR  BUILT-IN,  BEAMS  85 

and,  inserting  this  value  in  the  first  equation,  we  have 
(97)  «1  =  JP^1  +  £1). 

Also,  inserting  this  value  of  R1  in  the  expression  for  the  moment, 
it  is  found  that 

/  QQ\  J^A      _     J>       1    2 

Similarly,  by  forming  the  expressions  for  the  total  deflection  of  the 
point  E  with  respect  to  the  tangent  at  A,  we  obtain  the  equations 

R  7s         P/3        P/27 
_^_  M2i         ±  ^        ri^  i2  _  n 


3  El      3  El      2  El 
(99) 

_MJ,       Rf_       PI? 

El      2  El      2  El 

and,  solving  these  equations  simultaneously  for  M2  and  R^  as  above, 
the  results  are 


(wo) 


If  AC  is  the  longer  of  the  two  segments,  the  maximum  deflection 
will  occur  somewhere  between  A  and  C.  Also,  since  the  tangent  at 
this  point  must  be  horizontal,  the  total  angular  deflection  from  one 
end,  say  A,  to  the  point  of  maximum  deflection  is  zero.  Let  x 
denote  the  distance  of  this  point  of  maximum  deflection  from  A. 
Then,  computing  the  total  angular  deflection  up  to  this  point  and 
equating  it  to  zero,  we  have 


_  _ 

'  EI  +  2EI~ 

2Jf, 

whence  x  =  —  —• 

si 

Inserting  the  values  just  obtained  for  Ml  and  7^,  this  becomes 

(102) 


86  RESISTANCE  OF  MATERIALS 

The  maximum  deflection  D  is  then  found  to  be 


§EI 

and,  inserting  in  this  expression  the  values  of 
obtained,  it  reduces  to 


and  #,  just 


(103) 


48.  Uniformly    loaded   beam   fixed  at    one    end.    In    this    case 
(Fig.  76)  the  deflection  of  the  end  A  with  respect  to  the  fixed 

end  B   consists    of    two    parts: 
that  due  to  the  reaction  R  is 

Rl* 


and  that  dne  to  the  total  uniform 
load  wl  is 


SJSI 

Since   the    ends   A   and   B   are 
assumed    to    be     at    the    same 
level,  the  total  deflection   of  A 
from  the  tangent  at  B  must  be  zero ;  that  is, 


FIG.  76 


3  El      8  El 
whence  the  reaction  at  the  unrestrained  end  is 

(104) 

O 

The  reaction  Rr  at  the  restrained  end  B  is  therefore 

5  tvl 


(105) 


=  ivl  —  R  = 


8 


The  maximum  moment,  which  in  this  case  occurs  at  the  fixed  end 
B,  is  then 


(106) 


M=Kl-  —  =  _ 


RESTRAINED,  OR  BUILT-IN,  BEAMS 


87 


At  the  point  where  the  maximum  deflection  occurs  the  tangent 
is  horizontal.  Let  the  distance  of  this  point  from  the  fixed  end 
B  be  denoted  by  x.  Then,  from  (40),  (49),  and  (51),  articles  37, 
38,  and  39,  the  condition  that  the  total  angular  deflection  for 
this  length  x  shall  be  zero  is 

MX       R'x*        wx8  _  n 

~~EI  +  YEI~^EI  = 

Inserting  in  this  expression  the  values  of  M  and  R1  obtained  above, 
and  solving  for  a?,  the  result  is 


(107) 


=  — .  (15  -  >/33)  =  .578  I. 


The  maximum  deflection  is  then  found  by  finding  the  total  deflection 
for  the  length  x  with  respect  to  the  tangent  at  B.  Hence,  from 
(38),  (47),  and  (50),  articles  37,  38,  and  39, 


D      =- 

2  El 


R'x* 


wx 


or,  inserting  the  values  of  M  and  A*', 


(108) 


wx," 
48JEI 


(3 12  -  lOto  + 


The  numerical  value  of 
the  deflection  is  most  easily 
found  by  first  calculating 
the  numerical  value  of  x 
and  then  substituting  in 
this  formula. 

49.  Beam  fixed  at  one  end 
and  bearing  concentrated  load 
at  center.  The  deflection  of 
the  end  A  (Fig.  77)  with 

respect  to  the  fixed  end  B  in  this  case  consists  of  two  parts 
(38),  article  37,  that  due  to  the  reaction  R  is 

,        Rl* 


Moment  diagram 
FIG.  77 


from 


88  RESISTANCE  OF  MATERIALS 

and  from  (44),  article  37,  that  due  to  the  load  P  is 


22 


3  El 


Since  A  and  B  are  assumed  to  be  at  the  same  level,  the  total 
deflection  of  the  end  A  with  respect  to  the  tangent  at  B  must  be 
zero.  Consequently, 

Rla          PI3  PI3 


3  El      24  El      16  El 
5 


=  0; 


whence 

(109)  1 
Therefore 

(110)  R'  =  P-R  =  ^P. 

The  maximum  moment,  which  in  this  case  occurs  at  B,  is  then 


(in)  M  =  RI-  P- 


16 

The  position  and  amount 
of  the  maximum  deflec- 
tion may  be  found  as  in 
the  preceding  article. 

50.  Beam  fixed  at  one 
end  and  bearing  a  concen- 
trated eccentric  load.  The 
deflection  of  the  end  A 
with  respect  to  the  fixed  end  B  (Fig.  78)  also  consists  of  two  parts 
in  this  case.  From  (38),  article  37,  that  due  to  the  reaction  R  is 

Rf 


FIG.  78 


and  from  (44),  article  37,  that  due  to  the  load  P  is 


3  El      2  El 


RESTRAINED,  OR  BUILT-IN,  BEAMS  89 

Since  the  supports  are  assumed  to  be  at  the  same  level,  the  total 
deflection  of  A  with  respect  to  the  tangent  at  B  is  zero.  Con- 
sequently, 


3  El      3  El      2  El 
whence 


(112)  «  =          (a^ 

Also 

F*7 

(us)         «'  =  f  -  R  =  |J  [2  i(*  +  g  _  i*\. 

The  moment  at  the  fixed  end  B  is  then 

(114)  M  B  =  «l  -  .PI,  =  -  ^5  (I  +  O, 

—  / 

and  the  moment  at  the  load  C  is 

(115)  MC  =  ^1  =  ^1(2  ?  +  ?,). 

—   I 

The  maximum  deflection  may  be  found  by  the  method  explained 
in  article  48. 

APPLICATIONS 

156.  One  end  of  a  beam  is  built  into  a  wall,  and  the  other  end  is  supported  at 
the  same  level  by  a  post  12  ft.  from  the  wall.    The  beam  carries  a  uniform  load  of 
100  Ib.  per  linear  foot.    Find  the  position  and  amount  of  the  maximum  moment 
and  also  of  the  maximum  deflection. 

157.  One  end  of  a  beam  is  built  into  a  wall,  and  the  other  end  rests  on  a  prop 
20ft.  from  the  wall  at  the  same  level.    The  beam  bears  a  concentrated  load  of 
1  ton  at  a  point  8  ft.  from  the  wall.   Find  the  position  and  amount  of  the  maximum 
moment  and  also  of  the  maximum  deflection. 

158.  A  cantilever  of  length  I  is  loaded  uniformly.    At  what  point  of  its  length 
should  a  prop  be  placed,  supporting  the  beam  at  the  same  level  as  the  fixed  end, 
in  order  to  reduce  the  bending  stress  as  much  as  possible,  and  what  proportion  of 
the  load  is  then  carried  by  the  prop  ? 

159.  A  20-in.  steel  I-beam  weighing  65  Ib./ft.  is  built  into  a  wall  at  one  end 
and  rests  on  a  support  20  ft.  from  the  wall  at  the  other  end.    A  load  of  25  tons 
rests  on  the  beam  at  a  point  distant  15  ft.  from  the  wall.    Find  the  reaction  of  the 
support  and  the  maximum  deflection. 

160.  A  beam  of  uniform  section  is  built  into  a  wall  at  one  end,  projecting  16  ft. 
from  the  face  of  the  wall,  and  rests  on  a  column  at  12  ft.  from  the  wall.    The 
beam  carries  a  uniform  load  of  5  tons  per  foot  run.    Find  the  load  on  the  column. 


90  RESISTANCE  OF  MATERIALS 

161.  A  beam  of  uniform  section  is  built  into  walls  at  both  ends,  the  distance 
between  walls  being  25  ft.    Two  concentrated  loads,  each  of  5  tons,  rest  on  the 
beam  at  points  5  ft.  from  each  wall.    Find  the  maximum  bending  moment  in  the 
beam,  and  also  the  position  of  zero  bending  moment. 

162.  A  continuous  beam  of  two  spans,  each  of  40ft.,  carries  a  uniform  load  of 
1  ton  per  foot  run.    Find  the  reactions  of  the  supports  by  the  method  of  article  48, 
and  also  the  maximum  moment  and  maximum  deflection. 

163.  A  continuous  beam  of  two  equal  spans  is  uniformly  loaded.  Find  the  bend- 
ing moment  over  the  middle  support  when  the  three  supports  are  at  the  same  level, 
and  also  when  the  middle  support  is  raised  or  lowered  an  amount  h. 

164.  A  uniform  beam  of  20  ft.  span  is  fixed  at  both  ends  and  carries  a  load  of 
4  tons  at  the  center  and  two  loads  of  3  tons  each  at  5  ft.  from  each  end.   Find  the 
maximum  moment  and  the  position  of  zero  moment. 

165.  A  beam  of  length  2  I  is  supported  at  the  center,  one  end  being  anchored 
down  to  a  fixed  abutment  and  the  other  end  carrying  a  concentrated  load  W. 
Neglecting  the  weight  of  the  beam,  find  the  deflection  of  the  free  end. 


SECTION  VIII 

COLUMNS  AND  STRUTS 

51.  Nature  of  compressive  stress.     When  a  prismatic   piece  of 
length  equal  to  several  times  its  breadth  is  subjected  to  axial  com- 
pression, it  is  called  a  column,  or  strut,  the  word  column  being 
used  to  designate  a  compression  member  placed  vertically  and  bear- 
ing a  static  load,  all  other  compression  members  being  called  struts. 

If  the  axis  of  a  column  or  strut  is  not  perfectly  straight,  or  if  the 
load  is  not  applied  exactly  at  the  centers  of  gravity  of  its  ends,  a 
bending  moment  is  produced  which  tends  to  make  the  column  deflect 
sideways,  or  "  buckle."  The  same  is  true  if  the  material  is  not  per- 
fectly homogeneous,  causing  certain  parts  to  yield  more  than  others. 
Such  lateral  deflection  increases  the  bending  moment  and  conse- 
quently increases  the  tendency  to  buckle.  A  compression  member  is 
therefore  in  a  different  condition  of  equilibrium  from  one  subjected 
to  tension,  for  in  the  latter  any  deviation  of  the  axis  from  a  straight 
line  tends  to  be  diminished  by  the  stress  instead  of  increased. 

The  oldest  theory  of  columns  is  due  to  Euler,  and  his  formula  is 
still  the  standard  for  comparison.  Euler's  theory,  however,  is  based 
upon  the  assumptions  that  the  column  is  perfectly  straight,  the 
material  perfectly  homogeneous,  and  the  load  exactly  centered  at 
the  ends  —  assumptions  which  are  never  exactly  realized.  For 
practical  purposes,  therefore,  it  has  been  found  necessary  to  modify 
Euler's  formula  in  such  a  way  as  to  bring  it  into  accord  with  the 
results  of  actual  experiments,  as  explained  in  the  following  articles. 

52.  Euler's  theory  of  long  columns.   Consider  a  long  column  sub- 
jected to  axial  loading,  and  assume  that  the  column  is  perfectly 
straight  and  homogeneous  and  that  the  load  is  applied  exactly  at 
the  centers  of  gravity  of  its  ends.   Assume  also  that  the  ends  of  the 
column  are  free  to  turn  about  their  centers  of  gravity,  as  would  be 
the  case,  for  example,  in  a  column  with  round  or  pivoted  ends. 

91 


92 


RESISTANCE  OF  MATERIALS 


Now  suppose  that  the  column  is  bent  sideways  by  a  lateral  force, 
and  let  P  be  the  axial  load  which  is  just  sufficient  to  cause  the 
column  to  retain  this  lateral  deflection  when  the  lateral  force  is 
removed.  Let  OX  and  OY  be  the  axes  of  X  and  Y  respectively 
(Fig.  79).  Then  it  can  be  shown  that  the  elastic  curve  OCX  is 
a  sine  curve.  For  simplicity,  however,  it  will  be  assumed  to  be  a 
parabola.  Since  the  deflection  at  any  point  C  is  the  lever  arm 
of  the  load  P,  the  moment  at  C  is  Py.  The  moment  at  any  point 
is  therefore  P  times  as  great  as  the  deflection  at  that  point,  and  con- 
sequently the  moment  diagram  will  also  be  a  parabola  (Fig.  80). 

p 


Moment 
Diagram 


FIG.  79 


FIG. 80 


Now  let  d  denote  the  maximum  deflection,  which  in  this  case  is 
at  the  center.  Then  the  maximum  ordiiiate  to  the  moment  diagram 
is  Pd.  Therefore,  from  article  17,  the  area  of  one  half  the  diagram 

2  I      Pdl 

is  A  =  ^  (Pd)  -  =  —— ,  and  the  distance  of  its  centroid  from  one 

5    I      5 1 
end  is  XQ  =  —  •-  =  - — .   Hence,  from  the  general  deflection  formula, 

the  deflection  at  the  center  will  be 


(116) 


J_  J_  (Pld\  5J  _ 

"El      *~  EI\  3  )lQ~ 


Canceling  the  common  factor  d  and  solving  for  P,  the  result  is 

9.6  E  I 


(117) 


48  ££ 

"" 


COLUMNS  AND  STRUTS  93 

If  the  elastic  curve  had  been  assumed  to  be  a  sine  curve  instead 
of  a  parabola,  the  result  would  have  been  the  well-known  equation 

ir^EI       9.87  El 
(118)  P  =  __  =  __, 

which  is  Euler's  formula  for  long  columns  in  its  standard  form. 

Under  the  load  P  given  by  this  formula  the  column  is  in  neutral 
equilibrium ;  that  is  to  say,  the  load  P  is  just  sufficient  to  cause  it 
to  retain  any  lateral  deflection  which  may  be  given  to  it.  For  this 
reason  P  is  called  the  critical  load.  If  the  load  is  less  than  this 
critical  value,  the  column  is  in  stable  equilibrium,  and  any  lateral 
deflection  will  disappear  when  its  cause  is  removed.  If  the  load 
exceeds  this  critical  value,  the  column  is  in  unstable 
equilibrium,  and  the  slightest  lateral  deflection  will 
rapidly  increase  until  rupture  occurs. 

53.  Effect  of  end  support.  The  above  deduction  of 
Euler's  formula  is  based  on  the  assumption  that  the 
ends  of  the  column  are  free  to  turn,  and  therefore 
formula  (118)  applies  only  to  long  columns  with  round 
or  pivoted  ends. 

If  the  ends  of  a  column  are  rigidly  fixed  against 
turning,  the  elastic  curve  has  two  points  of  inflection, 
say  B  and  D  (Fig.  81).  From  symmetry,  the  tangent 
to  the  elastic  curve  at  the  center  C  must  be  parallel  to 
the  original  position  of  the  axis  of  the  column  AE,  and 
therefore  the  portion  AB  of  the  elastic  curve  must  be  symmetrical 
with  BC,  and  CD  with  DE.  Consequently,  the  points  of  inflection, 
B  and  D,  occur  at  one  fourth  the  length  of  the  column  from  either 
end.  The  critical  load  for  a  column  with  fixed  ends  is  therefore 
the  same  as  for  a  column  with  free  ends  of  half  the  length ;  whence, 
for  fixed  ends,  Euler's  formula  becomes 

(119) 

Columns  with  flat  ends,  fixed  against  lateral  movement,  are 
usually  regarded  as  coming  under  formula  (119),  the  terms  fixed 
ends  and  flat  ends  being  used  interchangeably. 


94 


RESISTANCE  OF  MATERIALS 


If  one  end  of  the  column  is  fixed  and  the  other  end  is  free  to 
turn,  the  elastic  curve  is  approximately  represented  by  the  line 
BCDE  in  Fig.  81.  Therefore  the  critical  load  in  this  case  is  ap- 
proximately the  same  as  for  a  column  with  both  ends  free,  of  length 
BCD,  that  is,  of  length  equal  to  |  BE  or  1 1 ;  whence,  for  a  column 
with  one  end  fixed  and  the  other  free,  Euler's  formula  becomes 


(120) 


P  = 


9  IT2 El 
4  I2 


approximately. 


One  end  I    Ends  fixed  n  direction;    One  end 


If  the  lower  end  is  fixed  in  direction  but  the  upper  end  is  entirely 
free  (that  is,  if  there  is  no  horizontal  reaction  to  prevent  it  from 
bending  out  sideways),  it  may  be  regarded  as  half  of  a  column 
with  round  or  pin 
ends  and  of  length 
2 1.  Consequently, 
in  this  case  Euler's 
formula  becomes 

7T2EI 


Round  ends; 
Position  fixed 
but  not  direc- 
tion. 


end 


The  general  expres- 
sion for  Euler's  formula 
is  then 


(122)      P  =  K 


TT^EI 

7 


Fm.  82 


where  the  constant  k  is  determined  by  the  way  in  which  the  ends 
of  the  column  are  supported.  The  values  of  k  corresponding  to 
various  end  conditions  are  given  in  Fig.  82. 

54.  Modification  of  Euler's  formula.  It  has  been  found  by  ex- 
periment that  Euler's  formula  applies  correctly  only  to  very  long 
columns,  and  that  for  short  columns  or  those  of  medium  length  it 
gives  a  value  of  P  considerably  too  large. 

Very  short  columns  or  blocks  fail  solely  by  crushing,  the  tend- 
ency to  buckle  in  such  cases  being  practically  zero.  Therefore,  if 
p  denotes  the  crushing  strength  of  the  material  and  A  the  area 


COLUMNS  AND   STRUTS  95 

of  a  cross  section,  the  breaking  load  for  a  very  short   column  is 
P=pA* 

For  columns  of  ordinary  length,  therefore,  the  load  P  must  lie 
somewhere  between  pA  and  the  value  given  by  Euler's  formula. 
Consequently,  to  obtain  a  general  formula  which  shall  apply  to 
columns  of  any  length,  it  is  only  necessary  to  express  a  continuous 

relation  between  pA  and  — —  •   Such  a  relation  is  furnished  by  the 
equation 

das)  r  = 

1+pA 


7T2EI 

For  when  I  =  0,  P=pA,  and  when  I  becomes  very  large,  P  approaches 

IT2  El 

the  value  —  —  •   Moreover,  for  intermediate  values  of  I  this  formula 
c 

gives  values  of  P  considerably  less  than  those  given  by  Euler's 
formula,  thus  agreeing  more  closely  with  experiment. 

55.  Rankine's  formula.  Although  the  above  modification  of 
Euler's  formula  is  an  improvement  on  the  latter,  it  does  not  yet 
agree  closely  enough  with  experiment  to  be  entirely  satisfactory. 
The  reason  for  the  discrepancy  between  the  results  given  by  this 
formula  and  those  obtained  from  actual  tests  is  that  the  assumptions 
upon  which  the  formula  is  based,  namely,  that  the  column  is  perfectly 
straight,  the  material  perfectly  homogeneous,  and  the  load  applied 
exactly  at  the  centers  of  gravity  of  the  ends,  are  never  actually 
realized  in  practice. 

To  obtain  a  more  accurate  formula,  two  empirical  constants  will 
be  introduced  into  equation  (123).  Thus,  for  fixed  ends,  let 


(124) 


where  /  and  g  are  arbitrary  constants  to  be  determined  by  experi- 
ment, and  t  is  the  least  radius  of  gyration  of  a  cross  section  of  the 
column.  This  formula  has  been  obtained  in  different  ways  by 

*  As  Euler's  formula  is  based  upon  the  assumption  that  the  column  is  of  sufficient 
length  to  buckle  sideways,  it  is  evident  a  priori  that  it  cannot  be  applied  to  very  short 
columns,  in  which  this  tendency  is  practically  zero.  Thus,  in  formula  (1 1 8) ,  as  I  approaches 
zero  P  approaches  infinity,  which  of  course  is  inadmissible. 


96  RESISTANCE  OF  MATERIALS 

Gordon,  Rankine,  Navier,  and  Schwarz.*  Among  German  writers 
it  is  known  as  Schwarz's  formula,  but  in  English  and  American 
textbooks  it  is  called  Rankine 's  formula. 

For  I  =  0,  P  =  gA,  and,  since  short  blocks  fail  by  crushing,  g  is 
therefore  the  ultimate  compressive  strength  of  the  material. 

For  different  methods  of  end  support  Rankine' s  formula  takes 
the  following  forms : 


(125)  —  = /7\2  Flat  endS 

1  _j_  /  /  _  ]  (fixed  in  direction) 


(126)  —  = 9         2  Round  ends 

•"        -^  _i_  A  f  I  _\  (direction  not  fixed) 

w 

(127)  -  = 9— r-2  Hinged  ends 

'  1   _J_  9  /"  /  _  \  (position  fixed,  but  not 

J    1  £  I  direction) 

(128)  —  = ry^  One  end  flat  and  the 

1+1.78/Y-)  other  round 

W 
56.  Values  of  the  empirical  constants  in  Rankine 's  formula.    The 

values  of  the  empirical  constants,  /  and  g,  in  Rankine's  formula 
have  been  experimentally  determined  by  Hodgkinson  and  Christie, 
with  the  following  results  : 

For  hard  steel,  g  =  69,000  lb./in.2,         /=  — !^. 

-i 
For  mild  steel,  g  =  48,000  lb./in.2,         /  = 


For  wrought  iron,         g  =  36,000  lb./in.2,         f  = 


30000 

1 
36000 


For  cast  iron,  g  =  80,000  lb./in.2,         /=  - 

For  timber,  g=    7,200  lb./in.2,        /= 


6400 
1 


3000 

*  Rankine's  formula  can  be  derived  independently  of  Euler's  formula  either  by 
assuming  that  the  elastic  curve  assumed  by  the  center  line  of  the  column  is  a  sinusoid 

or  by  assuming  that  the  maximum  lateral  deflection  D  at  the  center  of  the  column  is 

li 

given  by  the  expression  D  =  n  — ,  where  I  is  the  length  of  the  column,  b  its  least  width, 
and  M  an  empirical  constant. 


COLUMNS  AND  STRUTS  97 

These  constants  were  determined  by  experiments  upon  columns  for 
which  20  <  -  <  200,  and  therefore  can  only  be  relied  upon  to 

t/ 

furnish  accurate  results  when  the  dimensions  of  the  column  lie 
within  these  limits. 

As  a  factor  of  safety  to  be  used  in  applying  the  formula,  Rankine 
recommended  10  for  timber,  4  for  iron  under  dead  load,  and  5  for 
iron  under  moving  load. 

57.  Johnson's  parabolic  formula.  From  the  manner  in  which 
equation  (123)  was  obtained  and  afterwards  modified  by  the  intro- 
duction of  the  empirical  constants  /  and  g,  it  is  clear  that  Rankine's 
formula  satisfies  the  requirements  for  very  long  or  very  short  col- 
umns, while  for  those  of  intermediate  length  it  gives  the  average 
values  of  experimental  results.  A  simple  formula  which  fulfills 
these  same  requirements  has  been  given  by  Professor  J.  B.  Johnson, 
and  is  called  Johnson's  parabolic  formula. 

If  equation  (124)  is  written 

£=      -         9 

A       * 


and  then  y  is  written  for  p,  and  x  for  -  ,  Rankine's  formula  becomes 

I/ 


1+/^2 

For  this  cubic  equation  Johnson  substituted  the  parabola 
(129)  y  =  8  -  ex\ 

in  which  x  and  y  have  the  same  meaning  as  above,  and  8  and  e  are 
empirical  constants.  The  constants  8  and  e  are  then  so  chosen  that 
the  vertex  of  this  parabola  is  at  the  elastic  limit  of  the  material 
on  the  axis  of  loads  (or  F-axis),  and  the  parabola  is  also  tangent 
to  Euler's  curve.  In  this  way  the  formula  is  made  to  satisfy  the 
theoretical  requirements  for  very  long  or  very  short  columns,  and 
for  those  of  intermediate  length  it  is  found  to  agree  closely  with 
experiment. 


98 


RESISTANCE  OF  MATERIALS 


For  different  materials  and  methods  of  end  support  Johnson's 
parabolic  formulas,  obtained  as  above,  are  as  follows : 


KIND  OF  COLUMN 

FORMULA 

LIMIT  FOR  USE 

Mild  steel 

Hinged  ends 

—  =  42,000  -  .97 

(l)2 

|<15° 

p 

/7\2 

7 

Flat  ends 

-  =  42,000  -  .62 

© 

-^190 

Wrought  iron 

Hinged  ends 

—  =  34,000  -  .67 
A. 

tT 

z  _ 

Flat  ends 

—  =  34,000  -  .43 

©' 

1^210 

Cast  iron 

p                        95 

/7\2 

I  _ 

Round  ends 

-  =  60,000-  - 
A                         4 

(9 

Flat  ends 

—  =  60,000  -      ^ 
A                          4 

w 

1^120 

6 

Timber  (flat  ends) 

White  pine 

—  =    2,500-    .6 
A 

tr 

-,^     60 

Short-leaf  yellow  pine 

—  =    3,300-    .7 

©• 

•1?    60 

t' 

Long-leaf  yellow  pine 

—  =    4,000-    .8 
A. 

©' 

p?    60 

White  oak 

—  =    3,500-    .8 

0' 

^60 

The  limit  for  use  in  each  case  is  the  value  of  x  I  =  -  J  at  the  point 
where  Johnson's  parabola  becomes  tangent  to  Euler's  curve.  For 
greater  values,  of  -  Euler's  formula  should  therefore  be  used. 

v 

A  graphical  representation  of  the  relation  between  Euler's  for- 
mula, Rankine's  formula,  J.  B.  Johnson's  parabolic  formula,  and 
T.  H.  Johnson's  straight-line  formula  (considered  in  the  next 
article)  is  given  in  Fig.  83  for  the  case  of  a  wrought-iron  column 
with  hinged  ends.t 


*  In  the  formulas  for  timber  tf  is  the  least  lateral  dimension  of  the  column. 
t  For  a  more  extensive  comparison  of  these  formulas  see  Johnson's  framed  Structures, 
8th  ed.,  1905,  pp.  159-171;  also  Trans.  Amer.  Soc.  Civ,  Eng.,  Vol.  XV,  pp.  518-536. 


COLUMNS  AND  STRUTS 


99 


58.  Johnson's  straight-line  formula.  By  means  of  an  exhaustive 
study  of  experimental  data  on  columns  Mr.  Thomas  H.  Johnson 
has  shown  that  for  columns  of  moderate  length  a  straight  line 
can  be  made  to  fit  the  plotted  results  of  column  tests  as  exactly 
as  a  curve.  He  has  therefore  proposed  the  formula 


(130) 


P  I 

—  =  V  —  <7  — 

A  t 


50  100  150  200 

FIG.  83.    Wrought-Iron  Column  (Pin  Ends) 


250 


300 


1,  Euler's  formula ;  2,  T.  H.  Johnson's  straight-line  formula ;  3,  J.  B.  Johnson's  parabolic 
formula;  4,  Rankine's  formula 

or,  in  the  notation  of  the  preceding  article, 
(131)  y  =  v-ax, 

in  which  v  and  cr  are  empirical  constants,  this  being  the  equation 
of  a  straight  line  tangent  to  Euler's  curve.  This  formula  has  the 
merit  of  great  simplicity,  the  only  objection  to  it  being  that  for 
short  columns  it  gives  a  value  of  P  in  excess  of  the  actual  break- 
ing load.  The  relation  of  this  formula  to  those  which  precede  is 
shown  in  Fig.  83. 


100  RESISTANCE  OF  MATERIALS 

The  constants  v  ando-  in  formula  (130)  are  connected  by  the  relation 

(132) 


_v_     I     4v 

~3\3n7r2^' 


where  for  fixed  ends  n  =  1,  for  free  ends  n  =  4,  and  for  one  end 
fixed  and  the  other  free  n  =  1.78. 

The  following  table  gives  the  special  forms  assumed  by  Johnson's 
straight-line  formula  for  various  materials  and  methods  of  end 
support :  * 


KIND  OF  COLUMN 

FORM  i 

'LA 

LIMIT  FOR  USE 

Hard  steel 

Flat  ends 

—  =  80,000 

-3371 

-  P  158.0 

A 

t 

i  ^ 

Hinged  ends 

-  =  80,000 

-4141 

-  P  129.0 

A 

t 

t  ^ 

Round  ends 

-  =  80,000 

-534  - 

-  ^    99.9 

A 

t 

t 

Mild  steel 

Flat  ends 

—  =  52,500 

-179! 

-  ^  195.1 

'A 

t 

t  ^ 

Hinged  ends 

—  =  52,500 

-220  - 

-  P  159.3 

A 

t 

t 

Round  ends 

—  =  52,500 

-284- 

-  ^  123.3 

A 

t 

t 

Wrought  iron 

Flat  ends 

—  =  42,000 

-  128  - 

ip  218.1 

A 

t 

i 

Hinged  ends 

—  =  42,000 

-157! 

7^178.1 

A 

t 

i 

Round  ends 

—  =  42,000 

-203  - 

-  P  138.0 

A 

t 

i  ^ 

Cast  iron 

Flat  ends 

—  -  80,000 

-438  - 

-  ^  121.6 

A 

t 

i  ^ 

Hinged  ends 

—  =  80,000 

-537! 

-  ^    99.3 

A 

i 

i 

Round  ends 

-  =  80,000 

-  693  - 

7<    77.0 

A 

t 

t 

Oak 

Flat  ends 

—  =    5,400 
A 

-    28  1 

^ 

-  ^  128.1 

*  Trans.  Amer.  Soc.  Civ.  Eng.,  1886,  p.  530. 


COLUMNS  AND  STRUTS 


'101 


The  limit  for  use  in  each  case  is  the  value  of  x  I  =  -  j  f or  the  point 

at  which  Johnson's  straight  line  becomes  tangent  to  Euler's  curve. 
59.  Cooper's  modification  of  Johnson's  straight-line  formula.    In 

his  standard  bridge  specifications  Theodore  Cooper  has  adopted 
Johnson's  straight-line  formulas,  modifying  them  by  the  introduc- 
tion of  a  factor  of  safety.  Thus,  for  medium  steel,  Cooper  specifies 
that  the  following  formulas  shall  be  used  in  calculating  the  safe 
load.  For  chords 


(133) 
For  posts 

(134) 


-  =    8,000  -  30  -  for  live-load  stresses, 
A  t 

P  I 

—  =  16,000  —  60  -  for  dead-load  stresses. 

A  t 


P  I 

—  =    7,000  —  40-  for  live-load  stresses, 
A  t 

P  I 

—  =  14,000  —  80  -  for  dead-load  stresses, 

jGL  6 

-  =  10,000  -  60  -  for  wind  stresses. 
A  t 


For  lateral  struts 


(135) 


'  P  I 

—  =    9,000  -  50  -  for  initial  stresses. 


By  initial  stress  in  the  last  formula  is  meant  the  stress  due  to 
the  adjustment  of  the  bridge  members  during  construction. 

60.  Eccentrically  loaded  columns.  In  a  column  that  carries  an 
eccentric  load  (for  example,  a  column  carrying  a  load  on  a  bracket 
or  the  post  of  a  crane)  there  is  a  definite  amount  of  bending  stress 
due  to  the  eccentricity  of  the  load  in  addition  to  the  column  stress. 
As  the  nature  of  column  stress  is  such  that  it  is  impossible  to  de- 
termine its  amount,  the  simplest  method  of  handling  a  problem  of 
this  kind  is  to  determine  its  relative  security  against  failure  as  a 
column  and  failure  by  bending.  That  is  to  say,  first  determine  its 
factor  of  safety  against  failure  as  a  column  under  the  given  column 
load.  Then  consider  it  as  a  beam  and  find  the  equivalent  bending 


102 


RESISTANCE  OF  MATERIALS 


P  =  A    52,500  -  179  -      the    factor    of    safety 
\  '/ 

against  column  failure  is 


moment  which  would  give  the  same  factor  of  safety.    Finally,  com- 
bine this  equivalent  bending  moment  with  that  due  to  the  eccentric 
load,  and  calculate  the  unit  stress  from  the  ordi- 
nary beam  formulas. 

To  illustrate  the  method,  suppose  that  a  col- 
umn 18  ft.  long  is  composed  of  two  12-in.  I-beams 
each  weighing  40  lb./ft.,  and  carries  a  column  load 
of  20  tons  at  its  upper  end  and  also  an  eccentric 
load  of  10  tons  with  eccentricity  2  ft.,  as  shown 
in  Fig.  84.  Assuming  that  the  column  has  flat 
ends,  and  using  Johnson's  straight-line  formula, 


1 


FIG.  84 


^(52,500-179- 

6 


=  2 (11.76)  (52,500 -179  (47.3))  =  . 
60,000 


60,000 

Now  consider  the  column  as  a  beam  and  find  the  equivalent  central 
load  K  corresponding  to  the  factor  of  safety  just  found,  namely, 
17.3.  The  maximum  moment  in  a  simple  beam  bearing  a  concen- 
trated load  K  at  the  center  is  M=—  •  Hence,  from  the  beam 

Kl      pi.  4  JP/ 


whence 


Assuming 


formula  M=—   we  have  —  = 

e  4         e  le 

the  ultimate  strength  of  the  material  to  be  60,000  lb./in.2,  we  have 


60,000  .,    ..    2 
P  =  ~^r  Win.2, 


7=2(245.9)  in.4, 
e  =  6  in., 


17.3 
Z  =  216  in., 

and,  inserting  these  values,  the  equivalent  load  K  is  found  to  be 
4  x  60,000  x  491.8 


17.3  x  216  x  6 


=  5220  Ib. 


Now  the  eccentric  load  P^  acting  parallel  to  the  axis  of  the  column, 
produces  the  same  bending  effect  as  a  horizontal  reaction  TTat  either 
end,  where  HI  =  P^d.  The  bending  moment  at  the  center,  due  to  a 

TTJ 

reaction  H  perpendicular  to  the  axis  of  the  beam,  is,  however,  —  • 

ft 


COLUMNS  AND   STRUTS  103 

Hence  the  total  equivalent  moment  at  the  center  now  becomes 
_Kl     Hl_Kl     P^d  __  5220  x  216      20,000  x  24 

T""T  =  :T"  ~Y:       ~T~  ~2~ 

=  521,880  in.-lb. 

Consequently,  the  maximum  unit  stress  in  the  member  becomes 
M     521,880 


which  corresponds  to  a  factor  of  safety  of  about  .9. 

If  this  factor  of  safety  is  larger  than  desired,  assume  a  smaller 
I-beam  and  repeat  the  calculations. 

A  method  substantially  equivalent  to  the  above  is  to  assume  that 
the  stress  in  a  column  is  represented  by  the  empirical  factor  in  the 
column  formula  used.  Thus,  for  a  short  block  the  actual  compressive 
stress  p  is  given  by  the  relation  P  =  pA,  whereas  in  the  column 

formula  used  above,  namely,  P  =  A  (  52,500  —  179  -  )  ,  the  stress  p  is 


replaced  by  the  empirical  factor  52,500  —  179-  •   Consequently,  the 
fraction  -, 

52,500-179- 

v 


where  uc  denotes  the  ultimate  compressive  strength  of  the  material, 
represents  the  reduction  in  strength  of  the  member  due  to  its  slim- 
ness  and  method  of  loading ;  or,  what  amounts  to  the  same  thing, 
the  equivalent  unit  stress  in  the  column  is 

(136)  ;       Pl  U~ 


^'52,500-179- 

T/ 

Applying  this  method  to  the  numerical  problem  given  above,  we 
have  A  =  23.52, 

-  =  47.3, 

u  _  60,000  =136 

52,500  -  179  pW- 179x47.3-    '     ' 


104  RESISTANCE  OF  MATERIALS 

Hence  the  equivalent  stress  in  the  column  is 

^5^x1.86  =  3470  Win-' 

Also,  the  bending  stress,  produced  by  the  eccentricity  of  the  load,  is 


Consequently,  by  this  method,  the  total  stress  in  the  column  is 
found  to  be  <  347Q  +  2928  =  6398  lb./in.2 

If  a  formula  of  the  Rankine-Gordon  type  is  used,  namely, 
P  9 


the  equivalent  stress  pe  in  the  column,  due  to  the  given  load  P,  is 


9 

where  uc  denotes  the  ultimate  compressive  strength  of  the  material, 
as  above. 

APPLICATIONS 

166.  A  solid,  round,  cast-iron  column  with  flat  ends  is  15ft.  long  and  Gin.  in 
diameter.    What  load  may  be  expected  to  cause  rupture  ? 

167.  A  square  wooden  post  12  ft.  long  is  required  to  support  a  load  of  15  tons. 
With  a  factor  of  safety  of  10,  what  must  be  the  size  of  the  post  ? 

168.  Two  8-in.  steel  I-beams,  weighing  25.25  Ib. /ft.,  are  joined  by  latticework 
to  form  a  column  25  ft.  long.    How  far  apart  must  the  beams  be  placed,  center  to 
center,  in  order  that  the  column  shall  be  of  equal  strength  to  resist  buckling  in 
either  axial  plane  ? 

169.  Four  medium  steel  angles,  5  x  3  x  f  in.,  have  their  3-in.  legs  riveted  to  a 
|-in.  plate  so  as  to  form  an  I-shaped  built  column.   How  wide  must  the  plate  be  in 
order  that  the  column  shall  be  of  equal  strength  to  resist  buckling  in  either  axial 
plane  ? 

170.  A  hollow  wrought-iron  column  with  flat  ends  is  20  ft.  long,  7  in.  internal 
diameter,  and  10  in.  external  diameter.  Calculate  its  ultimate  strength  by  Rankine's 
and  Johnson's  formulas  and  compare  the  results. 

171.  Compute  the  ultimate  strength  of  the  built  column  in  problem  168  by 
Rankine's  and  by  Johnson's  formulas  and  compare  the  results. 

172.  Compute  the  ultimate  strength  of  the  column  in  problem  169  by  Rankine's 
and  by  Johnson's  straight-line  formulas  and  compare  the  results. 


COLUMNS  AND  STRUTS 


105 


FIG. 85 


173.  A  column  18  ft.  long  is  formed  by  joining  the  legs  of  two  10-in.  steel  chan- 
nels, weighing  30  lb./ft.,  by  two  plates  each  10  in.  wide  and  ^  in.  thick,  as  shown 
in  Fig.  85.  Find  the  safe  load  for  this  column  by  Johnson's  straight-line  formula, 
^_^  _~  using  a  factor  of  safety  of  4. 

174.  A  wrought-iron  pipe  10  ft.  long,  and  of  inter- 
nal and  external  diameter  3  in.  and  4  in.  respectively, 
bears  a  load  of  7  tons.    What  is  the  factor  of  safety  ? 

175.  What  must  be  the  size  of  a  square  steel  strut 
8  ft.  long,  to  transmit  a  load  of  5  tons  with  safety  ? 

176.  Design  a  column  16  ft.  long  to  be  formed  of 
two  channels  joined  by  two  plates  and  to  support  a 
load  of  20  tons  with  safety. 

177.  Using  Cooper's  formula  for  live  load,  design 
the  inclined  end  post  of  a  bridge  which  is  25  ft.  long 
and  bears  a  load  of  30  tons,  the  end  post  to  be  com- 
posed of  four  angles,  a  top  plate,  and  two  side  plates. 

178.  A  strut  16  ft.  long,  fixed  rigidly  at  both  ends, 

is  needed  to  support  a  load  of  80,000  Ib.    It  is  to  be  composed  of  two  pairs  of  angles 
united  with  a  single  line  of  1-in.  lattice  bars  along  the  central  plane.    Determine 
the  size  of  the  angles  for  a  factor  of  safety  of  5.   (Note  that  the  angles  must  be 
spread  I2'm.  to  admit  the  latticing.) 

179.  For  short  posts  or  struts,  such  as  are  ordinarily  used  in  building  construc- 
tion, it  is  customary  to  figure  the  safe  load  as  12,000  lb./in.2  of  cross-section  area 

for  lengths  up  to  90  times  the  radius  of  gyration  ;  that  is,  for  -  =  90.  To  what  factor 

of  safety  does  this  correspond,  by  Johnson's  straight-line  formula  ? 

180.  The  posts  used  to  support  a  girder  in  a  building  are  8  in.  x  8  in.  timbers 
8  ft.  long.    Find  the  diameter  of  a  solid  cast-iron  column  of  equal  strength. 

If  a  wrought-iron  pipe  4  in.  in  external  diameter  is  used,  what  must  be  its  thick- 
ness to  be  equally  safe  ? 

181 .  At  what  ratio  of  diameter  to  length  would 
a  round  mild-steel  strut  have  the  same  tendency 
to  crush  as  to  buckle  ? 

182.  A  load  of  100  tons  is  carried  jointly  by 
three  cast-iron  columns  20  ft.  long.  What  saving 
in  material   will  be  effected  by  using  a  single 
column  instead  of  three,  the  factor  of  safety  to 
be  15  in  both  cases  ? 

183.  Determine  the  proper  size   of  a  hard- 
steel  piston  rod  48  in.  long  for  a  piston  18  in. 
in  diameter  and  a  steam  pressure  of  80  lb./in.2 
Consult  table  for  proper  factor  of  safety. 

184.  The  side  rod  of  a  locomotive  is  9ft.  long  between  centers,  4  in.  deep,  and 
2  in.  wide.    The  estimated  thrust  in  the  rod  is  12  tons,  and  the  transverse  inertia 
and  gravity  load  20  Ib.  per  inch  of  length.    Determine  the  factor  of  safety. 

185.  The  vertical  post  of  a  crane  (Fig.  86)  is  to  be  made  of  a  single  I-beam.  The 
post  is  pivoted  at  both  ends  so  as  to  revolve  about  its  axis.    Find  the  size  of  I-beam 
required  for  factor  of  safety  of  4  and  for  dimensions  and  loading,  as  shown. 


8  Tons 


FIG.  86 


SECTION  IX 

TORSION 

61.  Maximum  stress  in  circular  shafts.  When  a  uniform  circular 
shaft,  such  as  is  shown  in  Fig.  87,  is  twisted  by  the  application  of 
moments  of  opposite  signs  to  its  ends,  every  straight  line  AB  paral- 
lel to  its  axis  is  deformed  into  part  of  a  helix,  or  screw  thread,  AC. 
The  strain  in  this  case  is  one  of  pure  shear  and  is  called  torsion. 


The  angle  (f>  is  called  the  angle  of  shear  and  is  proportional  to  the 
radius  BD  of  the  shaft.  The  angle  6  is  called  the  angle  of  twist 
and  is  proportional  to  the  length  AB  of  the  shaft. 

Consider  a  section  of  length  A#  cut  from  a  circular  shaft  by 
planes  perpendicular  to  its  axis  (Fig.  87).  Let  A0  denote  the  angle 
of  twist  for  this  section.  Then,  since  the  angle  of  twist  is  propor- 
tional to  the  length  of  the  shaft,  A0  :  6  =  A#  :  I  ;  whence 


Also,  if  <f)  and  A0  are  expressed  in  circular  measure, 
BC=$'AB  =  $kx,     and     BC  =  A<9  -  BD= 

Therefore  <f>  =  —  —  —  r--    From  Hooke's  law,  -±=  G.   Hence 
A#         I  <> 

(!37)  ,-«* 


Therefore  q  is  proportional  to  r;  that  is  to  say,  the  unit  shear  is 
proportional  to  its  distance  from  the  center,  being  zero  at  the 
center  and  attaining  its  maximum  value  at  the  circumference. 

106 


TOKSION  107 

If  q'  denotes  the  intensity  of  the  shear  at  the  circumference,  and 
a  denotes  the  radius  of  the  shaft,  then  the  shear  q  at  a  distance  r 
from  the  center  is  given  by  the  formula 


Now  if  q  denotes  the  intensity  of  the  shear  on  any  element  of 
area  A^4,  the  total  force  acting  on  this  element  is  q&A,  and  its  mo- 
ment with  respect  to  the  center  is  qAAr.  Therefore  the  total  internal 
moment  of  resistance  is  ^qkAr,  where  the  summation  extends  over 
the  entire  cross  section  ;  and  since  this  must  be  equal  to  the  exter- 
nal twisting  moment  Mt,  we  have 


Inserting  for  q  its  value  in  terms  of  the  radius,  q  =  —  ,  this  becomes 


or,  since  by  definition  ^r2AJ  =  Iv,  the  polar  moment  of  inertia  of 
the  cross  section,  . 

Mt  =  &. 

a 

For  a  solid  circular  shaft  of  diameter  7),  Ip  —  and   a  —  —  ; 

consequently, 


(138)  q'  -- 

±p 

For  a  hollow  circular  shaft  of  external  diameter  D  and  internal 
diameter  d,  Ip  =  — —  (D4  —  c?4)  and  a  =  —  ;  hence 

(139)  *'  =  =4^nr 


62.  Angle  of  twist  in  circular  shafts.    From  equation  (137), 

e-JL-JL. 

Gr      Ga 
Therefore,  for  a  solid  circular  shaft,  from  equation  (138), 

(140)  0 


108  RESISTANCE  OF  MATERIALS 

and  for  a  hollow  circular  shaft,  from  equation  (139), 

.  32  Mtl 

(141)  0  = 


If  Mt  is  known  and  6  can  be  measured,  equations  (140)  and  (141) 
can  be  used  for  determining  G.  If  G  is  known  and  6  measured,  these 
equations  can  be  used  for  finding  Mt ;  in  this  way  the  horse  power 
can  be  determined  from  the  angle  of  twist. 

63.  Power  transmitted  by  circular  shafts.  Let  H  denote  the 
number  of  horse  power  being  transmitted  by  a  circular  shaft,  n  its 
speed  in  revolutions  per  minute  (R.P.M.),  and  Mt  the  torque,  or 
twisting  moment,  acting  on  it,  expressed  in  inch-pounds.  Then,  since 
the  angular  displacement  of  Mt  in  one  minute  is  2  ?m,  the  work 
done  by  the  torque  in  one  minute  is  2  irnMt.  Also,  since  one  horse 
power  =  33,000  ft.-lb./min.  =  396,000  in.-lb./min.,  the  total  work 
done  by  the  shaft  in  one  minute  is  396,000  H.  Therefore 

2  irnMt  =  396,000  H\ 
whence 

(148)  Mt  =  «****»*  =  63,030  £  in.-,b. 

277™  n 

Therefore,'  if  it  is  required  to  find  the  diameter  D  of  a  solid  circular 
shaft  which  shall  transmit  a  given  horse  power  H  with  safety,  then, 
from  equation  (138), 

,_  16  Mt  _  321,000  H  m 
~~ 


whence 

(143)  J>  =  68.5    ^H 

As  safe  values  for  the  maximum  unit  shear  q',  Ewing  recom- 
mends 9000  lb./in.2  for  wrought  iron,  13,500  lb./in.2  for  steel,  and 
4500  lb./in.2  for  cast  iron.  Inserting  these  values  of  q'  in  formula 
(143),  it  becomes 

(144)  D  = 

where  for  steel  n  =  2.88,  for  wrought  iron  p  =  3.29,  and  for  cast 
iron  a  =  4.15. 


TORSION  109 

Expressed-  in  kilowatts  instead  of  horse  power,  this  formula 
becomes 

(145)  Z>  = 

where  for  steel  p  =  3.175,  for  wrought  iron  p  =  3.627,  and  for  cast 
iron  p  =  4.576. 

64.  Combined  bending  and  torsion.  In  many  cases  shafts  are 
subjected  to  combined  bending  and  torsion,  as,  for  instance,  when 
a  shaft  transmits  power  by  means  of  one  or  more  cranks  or  pulleys. 
In  this  case  the  bending  moment  Mb  at  any  point  of  the  shaft  pro- 
duces a  normal  stress  p  in  accordance  with  equation  (33),  article  34, 
that  is, 


and  the  torque  Mt  produces  a  shearing  stress  q  given  by  (138), 
article  61,  namely, 


In  more  advanced  works  on  the  strength  of  materials,  however,  it 
is  shown  that  the  maximum  and  minimum  normal  and  shearing 
stresses,  resulting  from  any  such  combination  as  the  above,  are 
given  by  the  relations  * 


(148)  pm&x 

min 

(149)  qm&x 

min 

Therefore,  inserting  in  these  expressions  the  values  of  p  and  q  as 
given  above,  the  maximum  and  minimum  normal  and  shearing 
stresses  in  terms  of  the  bending  and  twisting  moments  are  found 
to  be 

(150)  p         —  pUp  =fc  "/Jffr  +  -MT 2)      (called  Rankine's  formula), 

16      / 

(151)  tfmax  =  ±  — ^5  VM^J  +  M?  (called  Guest's  formula). 

min 


*  Slocum  and  Hancock,  Strength  of  Materials,  Revised  Edition,  pp.  24-25. 


110  RESISTANCE  OF  MATERIALS 

Note  that  Rankine's  formula  gives  the  principal  normal  stresses,  that 
is,  tension  or  compression,  whereas  Guest's  formula  gives  shear.  Since 
the  ultimate  strength  in  tension  or  compression  is  usually  different 
from  that  in  shear,*  in  designing  circular  shafts  carrying  combined 
stress  both  formulas  should  be  tried  with  the  same  working  stress  (or 
factor  of  safety),  and  the  one  used  which  gives  the  larger  dimensions. 
65.  Resilience  of  circular  shafts.  In  article  7  the  resilience  of  a 
body  was  denned  as  the  internal  work  of  deformation.  For  a  solid 
circular  shaft  this  internal  work  is 


where  Mt  is  the  external  twisting  moment  and  6  is  the  angle  of  twist. 
From  equation  (137),          6  =  -%-  =  ^- , 

(jrT          (jrCL 

and  from  equation  (138), 

Therefore  the  total  resilience  of  the  shaft  is 


and  consequently  the  mean  resilience  per  unit  of  volume  is 

0*3)  ,.-$-& 

66.  Non-circular  shafts.  The  above  investigation  of  the  distribu- 
tion and  intensity  of  torsional  stress  applies  only  to  shafts  of  circular 
section.  For  other  forms  of  cross  section  the  results  are  entirely 
different,  each  form  having  its  own  peculiar  distribution  of  stress. 

For  any  form  of  cross  section  whatever,  the  stress  at  the  boundary 
must  be  tangential,  for  if  the  stress  is  not  tangential,  it  can  be 
resolved  into  two  components,  one  tangential  and  the  other  normal 
to  the  boundary  ;  but  a  normal  component  would  necessitate  forces 
parallel  to  the  axis  of  the  shaft,  which  are  excluded  by  hypothesis. 

Since  the  stress  at  the  boundary  must  be  tangential,  the  circular 
section  is  the  only  one  for  which  the  stress  is  perpendicular  to  a 
radius  vector.  Therefore  the  circular  section  is  the  only  one  to 

*  The  shearing  strength  of  ductile  materials,  both  at  the  elastic  limit  and  at  the  ulti- 
mate stress,  is  about  four  fifths  of  their  tensile  strength  at  these  points. 


TORSION  111 

which  the  above  development  applies,  and  consequently  is  the  only 
form  of  cross  section  for  which  Bernoulli's  assumption  holds  true. 
That  is  to  say,  the  circular  section  is  the  only  form  of  cross  section 
which  remains  plane  under  a  torsional  strain. 

The  subject  of  the  distribution  of  stress  in  non-circular  shafts  has 
been  investigated  by  St.  Venant,  and  the  results  of  his  investigations 
are  summarized  below  (articles  67~70). 

67.  Elliptical  shaft.  For  a  shaft  the  cross  section  of  which  is  an 
ellipse  of  semi-axes  a  and  £>,  the  maximum  stress  occurs  at  the  ends 
of  the  minor  axis  instead  of  at  the  ends  of  the  major  axis,  as  might 
be  expected.  The  unit  stress  at  the  ends  of  the  minor  axis  is  given 
by  the  formula 

(154) 


and  the  angle  of  twist  per  unit  of  length  is 

9     Mt(n*  +  &2) 
81  = 


The  total  angle  of  twist  for  an  elliptical  shaft  of  length  I  is  therefore 


68.  Rectangular  and  square  shafts.  For  a  shaft  of  rectangular 
cross  section  the  maximum  stress  occurs  at  the  centers  of  the  longer 
sides,  its  value  at  these  points  being 

(157)  ?max  . 

lib  VV  +  b2 

in  which  h  is  the  longer  and  b  the  shorter  side  of  the  rectangle.   The 
angle  of  twist  per  unit  of  length  is,  in  this  case, 


For  a  square  shaft  of  side  b  these  formulas  become 

(159) 
and 
(160) 


112  RESISTANCE  OF  MATERIALS 

The  value  of  q  for  a  square  shaft  found  from  this  equation  is 

Mr 
about  15  per  cent  greater  than  if  the  formula  q  =  —  were  used,  and 

p 
the  torsional  rigidity  is  about  .88  of  the  torsional  rigidity  of  a 

circular  shaft  of  equal  sectional  area. 

69.  Triangular  shafts.    For  a  shaft  whose  cross  section  is  an 
equilateral  triangle  of  side  <?, 

(161) 


c 
and  the  angle  of  twist  per  unit  of  length  is 


The  torsional  rigidity  of  a  triangular  shaft  is  therefore  .73  of  the 
torsional  rigidity  of  a  circular  shaft  of  equal  sectional  area. 

70.  Angle  of  twist  for  shafts  in  general.  The  formula  for  the 
angle  of  twist  per  unit  of  length  for  circular  and  elliptical  shafts 
can  be  written 


(163) 


I 


G        A* 


in  which  Ip  is  the  polar  moment  of  inertia  of  a  cross  section  about 
its  center,  and  A  is  the  area  of  the  cross  section.  This  formula  is 
rigorously  true  for  circular  and  elliptical  shafts,  and  St.  Venant  has 
shown  that  it  is  approximately  true  whatever  the  form  of  cross 
section. 

APPLICATIONS 

186.  A  steel  wire  20  in.  long  and  .182  in.  in  diameter  is  twisted  by  a  moment 
of  20  in.-lb.  The  angle  of  twist  is  then  measured  and  found  to  be  6  —  18°  3r.  What 
is  the  value  of  G  determined  from  this  experiment  ? 

Solution.   From  equation  (140),  article  62, 

where,  in  the  present  case,  using  pound  and  inch  units,  Mt  =  20,  I  =  20,  D  =  182, 
and  6  =  18°31/  =  .3232  radians.  Substituting  these  numerical  values,  the  result  is 

G  =  11,490,000  lb./in.2 

187.  A  steel  shaft  5  in.  in  diameter  is  driven  by  a  crank  of  12-in.  throw,  the 
maximum  thrust  on  the  crank  being  10  tons.   If  the  outer  edge  of  the  shaft-bearing 
is  11  in.  from  the  center  of  the  crank  pin,  what  is  the  stress  in  the  shaft  at  this  point  ? 


TOESION 


113 


Solution.    Referring  to  Fig.  88,  the 
dimensions  in  the  present  case  are         ^—i 

dj  =  12  in.,        d2  =  11  in.  S 

Consequently, 

Mt=lQ-  2000  •  12  =  240,000  in.-lb. 
Mb  =  10  .  2000  .  11  =  220,000  in.-lb. 

Therefore,  from  equation  (150),  article  64, 


=  _  (220,000  +  V220,0002  +  240,000 2) 

7T  5 

=  22,200  lb./in.2, 

and  similarly,  from  equation  (151), 
16  / 


FIG.  88 


188.  If  P  and  Q  denote  the  unit  stresses  at  the  elastic  limits  of  a  material  in 

p 
tension  and  shear  respectively,  show  that  when  —  <  1  the  material  will  fail  in  ten- 

P  ^ 

sion,  whereas  when  —  >  1  it  will  fail  in  shear,  when  subjected  to  combined  bending 

and  torsion,  irrespective  of  the  relative  values  of  the  bending  and  twisting  moments. 
Solution.    Combining  Rankine's  and  Guest's  formulas,  we  have 

16Jf6 


p  -  q  = 


P 


Consequently,  if  the  bending  moment  is  zero,  p'  =  q',  or  —  =  1,  whereas  if  it  is  not 

q' 

zero,  p'  >  q'.   Similarly,  if  the  twisting  moment  is  zero,  —  =  2. 

q' 

Now  let  Ft  and  F8  denote  the  factors  of  safety  in  tension  and  shear  respectively. 
Then 


Ft 
Fs 


Since 


/  T)  Jjl 

,  the  fraction  —  ^=1.    Consequently,  if  —  <1  also,  then  —  <1;  that 

P'  Q  F8 

is,  Ft  <  Fs,  and  the  material  is  weaker  in  tension  than  in  shear.  The  second  part 
of  the  theorem  is  proved  in  a  similar  manner. 

For  a  complete  discussion  of  this  question  see  article  by  A.  L.  Jenkins,  En- 
gineering (London,  November  12,  1909),  pp.  637-639. 

189.  Three  pulleys  of  radii  8,  4,  and  6  in.  respectively  are  keyed  on  a  shaft  as 
shown  in  Fig.  89.  Pulley  No.  1  is  the  driving  pulley  and  transmits  30  H.P.  to  the 
shaft,  of  which  amount  10  H.P.  is  taken  off  from  pulley  No.  2  and  the  remaining 
20  H.P.  from  pulley  No.  3.  The  speed  is  50  R.P.M.,  the  belts  are  all  parallel,  and 
the  tension  in  the  slack  side  of  each  belt  is  assumed  to  be  one  half  the  tension  in  the 
tight  side.  Find  the  required  size  of  the  shaft  for  a  working  stress  of  12,000  lb./in.2 
in  tension  and  9000  lb./in.2  in  shear. 


114 


RESISTANCE  OF  MATERIALS 


Solution.  The  first  step  is  to  find  the  tensions  in  the  belts.  Since  power  is  the 
rate  of  doing  work,  and  1  H.P.  =  550  ft.-lb./sec.,  the  formula  for  power  may  be 
written 

xi  Fv 

Horse  power  = » 

55O 


k— — S1--^- -?-' 


FIG.  89 

where  F  denotes  the  effective  force,  or  difference  in  tension  in  the  two  sides  of  the 
belt,  expressed  in  pounds,  and  v  is  the  belt  speed  in  ft.  /sec.  Hence,  for  the  pulley 
of  8  in.  radius  transmitting  30  H.P.,  we  have 


60  .  550 

whence  F  =  4730  Ib.    Since  by  assumption  the  tension  on  the  tight  side  of  the  belt 
is  twice  that  on  the  slack  side,  their  values  are 

Tension  on  tight  side  =  9460  Ib. 
Tension  on  slack  side  =  4730  Ib. 

The  belt  tensions  for  the  other 
pulleys  are  calculated  in  a  simi- 
lar manner,  the  results  being 
indicated  on  Fig.  89. 

Considering  the  shaft  as  a 
beam,  the  load  at  each  pulley 
is  equal  to  the  sum  of  the  belt 
tensions  for  that  pulley,  as 
shown  in  Fig.  90.  The  reac- 
tions of  the  bearings  and  the 
bending-moment  diagram  are 
next  obtained,  the  results  being 
given  in  Fig.  90. 

The  maximum  bending  and  twisting  moments  thus  occur  at  pulley  No.  1,  their 

numerical  values  being 

Mb  =  364,230  in.-lb., 
Mt  =  37,840  in.-lb. 


Moment  diagram 
FIG.  90 


TOKSION  115 

Therefore,  substituting  in  equation  (150),  we  have 


whence 

D  =  6.725  in. 

and  similarly,  from  equation  (151), 

9000  =  —  V364,2302  +  37,840 2  ; 


whence 

D  =  5.842  in. 

The  proper  diameter  for  the  shaft  is  then  the  larger  of  these  two  values,  say  6f  in. 

190.  If  the  angle  of  twist  for  the  wire  in  problem  186  is  6  =  40°,  how  great  is 
the  torsional  moment  acting  on  the  wire  ? 

191.  Compare  the  angle  of  twist  given  by  St.  Venant's  general  formula  with 
the  values  given  by  the  special  formulas  in  articles  67,  68,  and  69. 

192.  A  steel  shaft  is  required  to  transmit  300  H.P.  at  a  speed  of  200  revolutions 
per  minute,  the  maximum  moment  being  40  per  cent  greater  than  the  average. 
Find  the  diameter  of  the  shaft. 

193.  Under  the  same  conditions  as  in  problem  192,  find  the  inside  diameter  of  a 
hollow  circular  shaft  whose  outside  diameter  is  6  in.    Also  compare  the  amount  of 
metal  in  the  solid  and  hollow  shafts. 

194.  The  semi-axes  of  the  cross  section  of  an  elliptical  shaft  are  3  in.  and  5  in. 
respectively.    What  is  the  diameter  of  a  circular  shaft  of  equal  strength  ? 

195.  An  oak  beam  6  in.  square  projects  4ft.  from  a  wall  and  is  acted  upon  at 
the  free  end  by  a  twisting  moment  of  25,000  ft.-lb.  How  great  is  the  angle  of  twist  ? 

196.  A  steel  shaft  10  ft.  long  between  centers  of  bearings  and  4  in.  in  diameter 
carries  a  pulley  14  in.  in  diameter  at  its  center.    If  the  driving  tension  in  the  belt 
is  250  lb.,  and  the  following  side  runs  slack,  what  is  the  maximum  stress  in  the 
shaft,  and  how  many  H.P.  is  it  transmitting  when  running  at  80  R.P.M.  ? 

197.  Find  the  required  diameter  of  a  solid  wrought-iron  circular  shaft  which  is 
required  to  transmit  150  H.P.  at  a  speed  of  60  R.P.M. 

198.  Find  the  angle  of  twist  in  problem  192. 

199.  Find  the  angle  of  twist  in  problem  193  and  compare  it  with  the  angle  of 
twist  for  the  solid  shaft  in  problem  198. 

200.  How  many  H.P.  can  a  hollow  circular  steel  shaft  of  15  in.  external  diam- 
eter and  11  in.  internal  diameter  transmit  at  a  speed  of  50  R.P.M.  if  the  maximum 
allowable  unit  stress  is  not  to  exceed  12,000  lb./in.2  ? 

201.  Find  the  diameter  of  a  structural-steel  engine  shaft  to  transmit  900  H.P. 
at  75  R.P.M.  with  a  factor  of  safety  of  10. 

202.  Find  the  factor  of  safety  for  a  wrought-iron  shaft  5  in.  in  diameter  which 
is  transmitting  60  H.P.  at  125  R.P.M. 

203.  A  structural-steel  shaft  is  60  ft.  long  and  is  required  to  transmit  500  H.P. 
at  90  R.P.M.  with  a  factor  of  safety  of  8,  and  to  be  of  sufficient  stiffness  so  that 
the  angle  of  to'rsion  shall  not  exceed  .5°  per  foot  of  length.    Find  its  diameter. 

204.  Under  the  same  conditions  as  in  problem  203,  find  the  size  of  a  hollow 
shaft  if  the  external  diameter  is  twice  the  internal. 


116 


RESISTANCE  OF  MATERIALS 


* 

1 

i 

1 

3 

6" 
J 

-  —  10- 


10 HH 


FIG.  91 


205.  A  hollow  wrought-iron  shaft  9  in.  in  external  diameter  and  2  in.  thick  is 
required  to  transmit  600  H.P.  with  a  factor  of  safety  of  10.    At  what  speed  should 

it  be  run  ? 

206.  A  horizontal  steel  shaft  4  in.  in 
diameter  and  10  ft.  long  between  centers 
of  bearings  carries  a  300-lb.  pulley  14  in. 
in  diameter  at  its  center.  The  belt  on  the 
pulley  has  a  tension  of  50  Ib.  on  the  slack 
side  and  175  Ib.  on  the  driving  side.  Find 
the  maximum  stress  in  the  shaft,  assum- 
ing that  the  belt  exerts  a  horizontal  pull 
on  the  shaft. 

207.  An  overhung  steel   crank,  like 

that  shown  in  Fig.  88,    carries  a  maximum  thrust  on  the  crank  pin  of  2  tons. 
Length  of  crank,  9  in. ;  distance  from  center  of  pin  to  center  of  bearing,  5  in. 
Determine  the  size  of  crank  and  shaft  for  a  factor  of  safety  of  5. 

208.  A    propeller    shaft 
9  in.  in    diameter  transmits 
1000  H.P.  at  90  R.P.M.    If 
the  thrust  on   the  screw  is 
12  tons,  determine  the  maxi- 
mum stress  in  the  shaft. 

209.  A   round    steel   bar 
2  in.  in  diameter,  supported 
at  points  4  ft.  apart,  deflects 
.029  in.  under  a  central  load 
of   300  Ib.  and    twists  1.62° 
in  a  length  of  2|  ft.  under 
a  twisting   moment  of  1500 
ft.-lb.   Find  E  and  G  for  the 
material. 

210.  A    steel    shaft  sub- 


\ 

\ 

! 

C 

I" 

1 

I 

1 

B 

*i 

! 

r 


FIG.  92 


E 


jected  to  combined  bending  and  torsion  has  an  elastic  limit  in  tension  of  64,600 
lb./in.2  and  an  elastic  limit  in  shear  of  29,170  lb./in.2   Show  that  Guest's  formula, 

I  rather  than  Rankine's,  applies  to 

p  this  material. 

j^,  211.  A  shaft  subjected  to  com- 

bined bending  and  twisting  is  made 
of  steel  for  which  the  elastic  limit 
in  tension  is  28,800  lb./in.2  and  the 
elastic  limit  in  shear  is  16,000  lb./in.2 
Show  that  if  the  bending  moment  is 
one  half  the  twisting  moment,  the 
shaft  will  be  weakest  in  shear, 
whereas  if  the  bending  moment  is 
twice  the  twisting  moment,  it  will  be  weakest  in  tension. 

212.  A  cast-iron  flanged  shaft  coupling  is  connected  by  eight  1^  in.  bolts,  the 
axis  of  each  bolt  being  6  in.  from  the  axis  of  the  shaft.    The  diameter  of  the 


—  *— 

1 

if 

i 

jf 

B' 

/^•A- 

*  8  ->+<  8  
FIG.  93 

TORSION  117 

shaft  is  5  in.    Find  the  shear  on  each  bolt  when  the  maximum  shearing  stress  in 
the  shaft  is  9000  lb./in.2 

213.  A  crank  shaft  revolves  in  bearings  at  A  and  B,  as  indicated  in  Fig.  91. 
The  cranks  are  in  the  same  plane,  and  the  crank-pin  pressures  P  and  Q  are  assumed 
to  act  at  right  angles  to  the  cranks.    If  P  =  2500  lb.,  find  Q  and  the  reactions  of 
the  bearings  at  A  and  B.   Find  also  the  maximum  stress  in  the  crank  pin  C  and 
draw  the  bending  moment  and  shear  diagrams. 

214.  A  crank  shaft  revolves  in  bearings  at  B   and  F  (Fig.  92)  and  carries 
two  cranks  C  and  E  in  the  same  plane.    The  shaft  transmits  a  pure  torque  at 
the  left  end,  and  the  crank-pin  pressures  are  assumed  to  act  perpendicular  to  the 
plane  of  the  cranks.    Find  the  stresses  in  the  cranks  C  and  E,  and  in  the  shaft  at 
B  and  Z),  and  draw  the  bending  moment  and  shear  diagrams. 

215.  A  crank  shaft  revolves  in  bearings  at  B  and  D  (Fig.  93),  the  planes  of 
the   two  cranks  being   90°  apart.    Taking   the  dimensions  given  in  the  figure, 
assume  P  —  3000  lb.  and  find  Q,  the  reactions  of  the  bearings,  and  the  stress  in  the 
crank  pin  C7. 


SECTION  X 


SPHERES  AND  CYLINDERS  UNDER  UNIFORM  PRESSURE 

71.  Hoop  stress.  When  a  hollow  sphere  or  cylinder  is  subjected 
to  uniform  pressure,  as  in  the  case  of  steam  boilers,  standpipes,  gas, 
water,  and  steam  pipes,  fire  tubes,  etc.,  the  effect  of  the  radial  pres- 
sure is  to  produce  stress  in  a  circumferential  direction,  called  hoop 
stress.  In  the  case  of  a  cylinder  closed  at  the  ends,  the  pressure  on 
the  ends  produces  longitudinal  stress  in  the  side  walls  in  addition 
to  the  hoop  stress. 

If  the  thickness  of  a  cylinder  or  sphere  is  small  compared  with 
its  diameter,  it  is  called  a  shell.  In  analyzing  the  stress  in  a  thin 
shell  subjected  to  uniform  pressure,  such  as  that  due  to  water, 
steam,  or  gas,  it  may  be  assumed  that  the  hoop  stress  is  distributed 
uniformly  over  any  cross  section  of  the  shell.  This  assumption  will 
be  made  in  what  follows. 

72.  Hoop  tension  in  hollow  sphere.  Consider 
a  spherical  shell  subjected  to  uniform  internal 
pressure,  and  suppose  that  the  shell  is  cut  into 
hemispheres  by  a-  diametral  plane  (Fig.  94). 
Then,  if  iv  denotes  the  pressure  per  unit  of 
area  within  the  shell,  the  resultant  force  act- 
ing on  either  hemisphere  is  P  =  — - — ?  where 


FIG.  94 


d  is  the  radius  of  the  sphere.  If  p  denotes  the  unit  tensile  stress  on 
the  circular  cross  section  of  the  shell,  the  total  stress  on  this  cross 
section  is  irdlip,  approximately,  where  h  is  the  thickness  of  the  shell. 

Consequently,  ^^ 

=  Trdhp ; 

whence 

wd 
(165)  p  =  —, 


which  gives  the  hoop  tension  in  terms  of  the  radial  pressure, 

118 


SPHERES  AND   CYLINDERS 


119 


73.  Hoop  tension  in  hollow  circular  cylinder.    In  the  case  of  a 
cylindrical  shell,  its  ends  hold  the  cylindrical  part  together  in  such 
a  way  as  to  relieve  the  hoop  tension  at  either  extremity.    Suppose, 
then,  that  the  portion  of  the  cylinder  considered  is  so  far  removed 

from  either  end  that  the  influence  of  the 
end  constraint  can  be  assumed  to  be  zero. 
Suppose  the  cylinder  cut  in  two  by  a 
plane  through  its  axis,  and  consider  a  sec- 
tion cut  out  of  either  half  cylinder  by  two 
planes  perpendicular  to  the  axis,  at  a  dis- 
tance apart  equal  to  c  (Fig.  95).  Then  the 
FlG  95  resultant  internal  pressure  P  on  the  strip 

under  consideration  is  P  =  cdw,  and  the 

resultant  hoop  tension  is  2  chp,  where  the  letters  have  the  same  mean- 
ing as  in  the  preceding  article.  Consequently,  cdw  =  2  clip  ;  whence 

(166)  P  =  ^' 

This  result  is  applicable  to  shells  under  both  inner  and  outer 
pressure,  if  p  is  taken  to  be  the  excess  of  the  internal  over  the 
external  pressure. 

74.  Longitudinal  stress  in  hollow 
circular  cylinder.    If  the   ends   of  a 
cylinder  are  fastened  to  the   cylin- 
drical   part,    the    internal    pressure 
against  the  ends  produces  longitudi- 
nal   stresses  in   the    side  walls.    In 
this   case    the    cylindrical    part    is 

subjected  both    to  hoop    tension   and   to   longitudinal  tension. 

To  find  the  amount  of  the  longitudinal  tension,  consider  a  cross 
section  of  the  cylinder  near  its  center,  where  the  influence  of  the 
end  restraints  can  be  assumed  to  be  zero  (Fig.  96).  Then  the  re- 

72 

sultant  pressure  on  either  end  is  P  =  —  — »  and  the  resultant  longi- 
tudinal stress  on  the  cross  section  is  Trdlip.  Therefore  — —  =  Trdhp ; 
whence 

(167)  p  = 


\A 


IB 

FIG.  96 


ivd 


120 


RESISTANCE  OF  MATERIALS 


This  is  the  same  formula  as  for  the  sphere,  which  was  to  be 
expected,  since  the  cross  section  is  the  same  in  both  cases. 

75.  Thick  cylinders.   Lame's  formulas.    Consider  a  thick  circular 
cylinder  of  external  radius  a  and  internal  radius  5,  subjected  to  either 

internal  or  external 
uniform  pressure, or 
to  both  simultane- 
ously, and  suppose 
that  a  section  is  cut 
out  of  the  cylinder  by 
two  planes  perpen- 
dicular to  the  axis 
at  a  unit  distance 
FIG.  97  apart  (Fig.  97). 

Now    consider  a 

thin  ring  of  the  material  anywhere  in  the  given  section,  of  external 
radius  re  and  internal  radius  r{.  Then,  under  the  strain,  re  will  become 


where  se  denotes  the  unit  deformation  of  the  fiber,  which  never 
exceeds  T-oVo  ^or  sa^e  working  stresses.    Similarly,  ri  will  become 


where  the  unit  deformation  «t.  is  also  very  small.  Since  any  safe 
strain  produces  no  appreciable  change  in  the  sectional  area  of  the 
thin  ring  here  considered,  by  equating  its  sectional  areas  before  and 
after  strain  we  have 


Canceling  out  the  common  factor  IT  and  reducing,  this  becomes 

»•.*(«?+  20  =  »•«•(<?  +  *O; 

or,  since  the  unit  deformations  se  and  «t.  are  very  small,  their  squares 
may  be  neglected  in  comparison  with  their  first  powers,  and  con- 
sequently this  expression  further  reduces  to 


SPHERES  AND  CYLINDERS  121 

Since  by  Hooke's  law  the  unit  stress  is  proportional  to  the  unit 
deformation  within  the  elastic  limit,  if  pe  denotes  the  unit  stress  on 
the  outside  fiber  of  the  thin  ring,  and  p{  on  the  inside  fiber,  then 


Pe          *e         r 

or  Pir?=per?. 

Hence,  if  ph  denotes  the  hoop  stress  on  any  element  of  the  thin  ring 
and  r  the  radius  of  this  element,  then 

(168)  Pn1^  —  constant,  say  C. 

It  should  be  noted  that  this  relation  applies  only  to  a  thin  ring  and 
not  to  the  thick  cylinder  as  a  whole.  It  may  be  used,  however,  to 
find  the  change  in  the  hoop  stress  corresponding  to  a  small  change 
in  the  radius,  that  is  to  say,  the  difference  in  the  hoop  stress  on  two 
adjacent  fibers,  as  explained  in  what  follows. 

Now  again  consider  a  thin  ring  of  the  material  and  let  its  inter- 
nal radius  be  r  and  its  thickness  Ar.  Also,  let  ph  denote  the  hoop 
stress  in  this  thin  ring,  pr  the  radial  stress  acting  on  its  inner  sur- 
face, and  pr  +  A/?r  the  radial  stress  acting  on  its  outer  surface.  Then 
the  difference  in  pressure  on  the  inside  and  outside  of  the  ring 
must  be  equal  to  the  total  force  holding  the  ring  together  ;  that  is, 

Or  +  Apr)  2  (r  +  Ar)  -  2  rpr  =  2pAr  5 

but,  since  Aft.  Ar  is  infinitesimal  in  comparison  with  the  other  terms, 
this  reduces  to 

(169)  Ph=Pr  +  r-j£. 

If  the  ends  of  the  cylinder  are  free  from  restraint,  or  if  the  cyl- 
inder is  subjected  to  a  uniform  longitudinal  stress,  the  longitudinal 
deformation  must  be  constant  throughout  the  cylinder.  But  the 
lateral  action  of  pr  and  ph  produce  longitudinal  deformation  in 

accordance  with  Poisson's   law  (article  8).    Thus,  if  —   denotes 

m 

Poisson's  ratio,  the  longitudinal  deformation  due  to  the  action  of 

pr  and  ph  is  .&-  +  -&-,  or  -  (pf+Ph)-   Therefore,  in  order  that 
mE 


122  RESISTANCE  OF  MATERIALS 

this  expression  may  be  constant,  pr  +  ph  must  be  constant.  Denoting 
this  constant  by  &,  we  have 

(170)  Pr  +  Ph  =  k. 

Now  eliminating  ph  between  equations  (168)  and  (170),  we  have 
(t-p,X  =  G 

As  the  radius  r  increases,  the  stress  pr  increases  or  decreases  ac- 
cording to  whether  the  constant  C  is  positive  or  negative  ;  that  is, 
whether  the  internal  pressure  is  greater  or  less  than  the  external. 
Since  the  sign  of  C  has  no  effect  on  the  result,  we  may  say  that  for 
a  point  at  a  distance  r  -f-  Ar  from  the  axis  the  radial  stress  is  of 
amount  pr  —  Apr,  such  that 

\k  -  (>r  +  A?,.)]  (r  +  Ar)2  =  C. 
Simplifying  this  expression,  it  becomes 

(k  -  pr)  r2  +  2  r  Ar  (k  -  pr)  -  r^pr  =  C, 
and  subtracting  from  it  the  original  relation,  namely, 


we  have  2  rAr  (Tc  —  pr)  —  r*&pr  =  0, 

Apr      ZrQc-pr)      Zph 
whence  -^  =  -       ~-^-  =  -^  ; 

Ar  r2  r 

C 
and,  since  ph  =  —  5  this  becomes 


Substituting  equation  (171)  in  equation  (169)  and  making  use 
of  equation  (170),  we  have 

2  C  2  C 

P*ssPr  +  -f  =  *-p*  +  -jri 

whence 


and  therefore,  from  (170), 


SPHERES  AND  CYLINDERS  123 

If,  therefore,  the  cylinder  is  subjected  to  a  uniform  internal  pres- 
sure of  amount  w{  per  unit  of  area,  and  also  to  a  uniform  external 
pressure  of  amount  we  per  unit  of  area,  then  pr  =  we  when  r  =  a, 
and  pr  =  tv{  when  r  =  b.  Substituting  these  simultaneous  values 
in  equation  (173), 

k      C  k      C 

w  =  ---  »          w.  =  ---  1 
e      2      a2  l      2      W 

a?b2  (we  —  w.)  k      wa2  —  wtf 

whence         c=      i        J.        —- 


Hence,  substituting  these  values  of   C  and  —  in  equations   (172) 
and  (173),  they  become 


Pr  = 


(^  -b2)r2 

(174) 

Ph  =  -^ 7T—  + 


(a2-b2)r2 

which  give  the  radial  and  hoop  stresses  in  a  thick  cylinder  subjected 
to  internal  and  external  pressure.  Equations  (174)  are  known  as 
Lame's  formulas. 

76.  Maximum  stress  in  thick  cylinder  under  uniform  internal 
pressure.  Consider  a  thick  circular  cylinder  which  is  subjected  only 
to  internal  pressure.  Then  we  =  0,  and  equations  (174)  become 


(175)       »=_!_(--—  1 
2        2     2 


w,b2 
p=  --  —^- 

r2        /  a2  —  b2 

Since  ph  is  negative,  the  hoop  stress  in  this  case  is  tension. 

Since  pr  and  ph  both  increase  as  r  decreases,  the  maximum  stress 
occurs  on  the  inner  surface  of  the  cylinder,  where  r  =  b  and  pr  =  w{. 
Hence 


Clearing;  the  latter  of  fractions,  we  have  —  =  —  -  -  5  whence  the 

b2     ph  -  w{ 

thickness  of  the  tube,  h  —  a  —  b,  is  given  by 


124  RESISTANCE  OF  MATERIALS 

77.  Bursting  pressure  for  thick  cylinder.  Let  ut  denote  the  ulti- 
mate tensile  strength  of  the  material  of  which  the  cylinder  is  com- 
posed. Then,  from  equation  (176),  the  maximum  allowable  internal 
pressure  wi  is  obtained  from  the  equation 


whence 

(178)  wi  =  "*\    ~  2     • 

Equation  (178)  gives  the  maximum  internal  pressure  u\  which  the 
cylinder  can  stand  without  bursting. 

78.  Maximum  stress  in  thick  cylinder  under  uniform  external 
pressure.  Consider  a  thick  circular  cylinder  subjected  only  to 
external  pressure.  In  this  case  w{  =  0  and  equations  (174)  become 

wa' 


Since  ph  is  positive,  the  hoop  stress  in  this  case  is  compression. 
For  a  point  on  the  inner  surface  of  the  cylinder  r  =  b,pr  =  Q,  and 

(179) 

79.  Comparison  of  formulas  for  the  strength  of  tubes  under  uni- 
form internal  pressure. 

I.  Thin   Cylinder  f-=.O23j.    From  article  73  the  formula  for 
the  hoop  (or  circumferential)  stress  in  a  thin  circular  cylinder  is 

(180)  Pft  =  g, 

and  from  article  74,  when  the  ends  of  the  cylinder  are  closed,  the 
longitudinal  stress  is 

(18!) 


SPHERES  AND  CYLINDERS  125 

The  actual  stress  in  a  thin  cylinder,  due  to  the  combination  of  these 
two  stresses  and  based  on  a  value  of  Poisson's  ratio  =  .3,  is  then 
found  to  be* 

(182)  1>=.425  — . 

II.  Thick  Cylinder.  Lame's  Formula.  In  article  76  the  maxi- 
mum stress  in  a  thick  cylinder  under  uniform  internal  pressure  is 
given  by  equation  (176)  in  terms  of  the  radii  a  and  b.  If  the  internal 
and  external  diameters  of  the  tube  are  denoted  by  d  and  D  respec- 
tively, then  d—  2£>,  D=  2 a,  and  the  formula  becomes 


(183)  p  = 


III.  Barloitfs  Formula.  This  formula,  which  is  widely  used 
because  of  its  simplicity,  assumes  that  the  area  of  cross  section  of 
the  tube  remains  constant  under  the  strain,  and  that  the  length 
of  the  tube  also  remains  unaltered.  As  neither  of  these  assump- 
tions is  correct,  the  formula  can  give  only  approximate  results. 
In  the  notation  previously  used  Barlow's  formula  is 

(184)  p 

It  is  therefore  of  the  same  form  as  the  formula  for  the  hoop  stress 
in  a  thin  cylinder,  except  that  it  is  expressed  in  terms  of  the  out- 
side diameter  D  inside  of  the  inside  diameter. 

From  the  results  of  their  experience  in  the  manufacture  and 
testing  of  tubes,  the  National  Tube  Company  asserts  that  for  any 

ratio  of  —  <  .3  Barlow's  formula  "  is  best  suited  for  all  ordinary 

calculations  pertaining  to  the  bursting  strength  of  commercial  tubes, 
pipes,  and  cylinders." 

For  certain  classes  of  seamless  tubes  and  cylinders,  however,  and 
for  critical  examination  of  welded  pipe,  where  the  least  thickness 
of  wall,  yield  point  of  the  material,  etc.  are  known  with  accuracy, 
and  close  results  are  desired,  they  recommend  that  the  following  for- 
mulas, due  to  Clavarino  and  Birnie,  be  used  rather  than  Barlow's. 

*  Slocum  and  Hancock,  Strength  of  Materials,  Revised  Edition,  p.  156. 


126  RESISTANCE  OF  MATERIALS 

IV.  Clavarino's  formula.  In  this  formula  each  particle  of  the 
tube  is  assumed  to  be  subjected  to  radial  stress,  hoop  stress,  and 
longitudinal  stress,  due  to  a  uniform  internal  pressure  acting  jointly 
on  the  tube  wall  and  its  closed  ends.  The  formula  also  involves 
Poisson's  ratio  of  lateral  contraction,  and  is  theoretically  correct, 
provided  the  maximum  stress  does  not  exceed  the  elastic  limit  of 
the  material.  Assuming  a  value  of  Poisson's  ratio  =  .3  and  using 
the  same  notation  as  above,  Clavarino's  formula  is 

- 
*• 

whence 


(186) 


V.  Birnies  formula.  This  formula  is  based  upon  the  same 
assumptions  as  Clavarino's,  except  that  the  longitudinal  stress  is 
assumed  to  be  zero.  Using  the  same  notation  as  before  and  assum- 
ing Poisson's  ratio  for  steel  to  be  .3,  Birnie's  formula  is 

7<l2) 


whence 
(188) 


10  1> 


80.  Thick  cylinders  built  up  of  concentric  tubes.  From  equations 
(174)  it  is  evident  that  in  a  thick  cylinder  subjected  to  internal 
pressure  the  stress  is  greatest  on  the  inside  of  the  cylinder  and 
decreases  toward  the  outside.  In  order  to  equalize  the  stress 
throughout  the  cylinder  and  thus  obtain  a  more  economical  use 
of  material,  the  device  used  consists  in  forming  the  cylinder  of 
several  concentric  tubes  and  producing  an  initial  compressive  stress 
on  the  inner  ones.  For  instance,  in  constructing  the  barrel  of  a 
cannon  or  the  cylinder  of  a  hydraulic  press  the  cylinder  is  built 
up  of  two  or  more  tubes.  The  outer  tubes  in  this  case  are  made  of 
somewhat  smaller  diameter  than  the  inner  tubes,  and  each  is 
heated  until  it  has  expanded  sufficiently  to  be  slipped  over  the  one 
next  smaller.  In  cooling,  the  metal  of  the  outer  tube  contracts, 
thus  producing  a  compressive  stress  in  the  inner  tube  and  a  tensile 


SPHERES  AND  CYLINDERS  127 

stress  in  the  outer  tube.  If,  then,  this  composite  tube  is  subjected 
to  internal  pressure,  the  first  effect  of  the  hoop  tension  thus  pro- 
duced is  to  relieve  the  initial  compressive  stress  in  the  inner  tube 
and  increase  that  in  the  outer  tube.  Thus  the  resultant  stress  in 
the  inner  tube  is  equal  to  the  difference  between  the  initial  stress 
and  that  due  to  the  internal  pressure,  whereas  the  resultant  stress 
in  the  outer  tube  is  equal  to  the  sum  of  these  two.  In  this  way  the 
strain  is  distributed  more  equally  throughout  the  cylinder.  It  is 
evident  that  the  greater  the  number  of  tubes  used  in  building 
up  the  cylinder,  the  more  nearly  can  the  strain  be  equalized. 

The  preceding  discussion  of  the  stress  in  thick  tubes  can  also  be 
applied  to  the  calculation  of  the  stress  in  a  rotating  disk.  For  ex- 
ample, a  grindstone  is  strained  in  precisely  the  same  way  as  a  thick 
tube  under  internal  pressure,  the  load  in  this  case  being  due  to 
centrifugal  force  instead  of  to  the  pressure  of  a  fluid  or  gas. 

81.  Practical  formulas  for  the  collapse  of  tubes  under  external 
pressure.  A  rigorous  analysis  of  the  stress  in  thin  tubes,  due  to 

external  pressure,  using  Poisson's  ratio  —  of  transverse  to  longi- 
tudinal deformation,  gives  the  formula  * 


=     *     (-} 

4fi-.i:>w 


or,  in  terms  of  the  diameter  D  =  2  «, 

w~    iU 

1        m2 

This  formula,  however,  is  based  on  the  assumptions  that  the  tube 
is  perfectly  symmetrical,  of  uniform  thickness,  and  of  homogeneous 
material  —  conditions  which  are  never  fully  realized  in  commercial 
tubes.  From  recent  experiments  on  the  collapse  of  tubes,  f  how- 
ever, it  is  now  possible  to  determine  the  practical  limitations  of 
this  formula  and  to  so  modify  it,  by  a  method  similar  to  that  by 

*Love,  Mathematical  Theory  of  Elasticity,  Vol.  II,  pp.  308-316. 
t  Carman,  "  Resist,  of  Tubes  to  Collapse,"  Univ.,  III.  Bull.,  Vol.  Ill,  No.  17 ;  Stewart, 
"Collap.  Press.  Lap- Welded  Steel  Tubes,"  Trans.  A.S.M.E.,  1906,  pp.  730-820. 


128  RESISTANCE  OF  MATERIALS 

which  the  Gordon-Rankine  column  formula  was  deduced  from 
Euler's  formula  (articles  54,  55),  as  to  obtain  a  rational  formula 
which  shall,  nevertheless,  conform  closely  to  experimental  results. 
By  determining  the  ellipticity,  or  deviation  from  roundness,  and 
the  variation  in  thickness  of  the  various  types  of  tubes  covered  by 
the  tests  mentioned  above,  it  is  found  that  by  introducing  empirical 
constants  the  rational  formulas  can  be  made  to  fit  experimental  re- 
sults as  closely  as  any  empirical  formulas,  with  the  advantage  of 
being  unlimited  in  their  range  of  application.*  The  formula  so 
obtained  is 

,1QQ,                                          2EC   /h\3  f for  thin  tubes 

(^ley;  w  = 


1--V~'  I      -^.023 

m?  [     D 

where  h  =  average  thickness  of  tube  in  inches, 

D  =  maximum  outside  diameter  in  inches, 

—  =  Poisson's  ratio  =  .3  for  steel, 

771 

C  =  .6  9  for  lap-welded  steel  boiler  flues, 
=  .76  for  cold-drawn  seamless  steel  flues, 
=  .78  for  drawn  seamless  brass  tubes. 

By  a  similar  procedure  for  thick  tubes  (—  >.023j  a  practical 

rational  formula  has  been  obtained  from  Lame's  formula  (article  75) 
for  this  case  also,  namely, 

I  v  [for  thick  tubes 

(      D" 

where  uc  =  ultimate  compressive  strength  of  the  material, 

7jT  =  .89  for  lap-welded  steel  boiler  flues. 

Only  one  value  of  K  is  given,  as  the  experiments  cited  were  all 
made  on  one  type  of  tube. 

The  correction  constants  C  and  K  include  corrections  both  for 
ellipticity,  or  flattening  of  the  tube,  and  for  variation  in  thickness. 

*  Slocum,  "  The  Collapse  of  Tubes  under  External  Pressure,"  Engineering.  London, 
January  8,  1909.   Also  abstract  of  same  article  in  Kent,  8th  ed.,  1910,  pp.  320-322. 


SPHERES  AND  CYLINDERS 


129 


Thus,  if  the  correction  for  ellipticity  is  denoted  by   Cl  and  the 
correction  for  variation  in  thickness  by  C2,  we  have 

_  minimum  outside  diameter 
1      maximum  outside  diameter 

_  minimum  thickness 
average  thickness 

and  the  correction  constants  C  and  K  are  therefore  denned  as 

c=c*c*, 


By  an  experimental  determination  of  C1  and  C2  the  formulas  can 
therefore  be  applied  to  any  given  type  of  tube. 

82.  Shrinkage  and  forced  fits.  In  machine  construction  shrink- 
age and  forced,  or  pressed,  fits  are  frequently  employed  for  connect- 
ing certain  parts,  such  as  crank  disk  and  shaft,  wheel  and  axle,  etc. 
To  make  such  a  connection  the  shaft  is  finished  slightly  larger  than 
the  hole  in  the  disk  or  ring  in  which  it  belongs.  The  shaft  is  then 
either  tapered  slightly  at  the  end  and  pressed  into  the  ring  cold,  or 
the  ring  is  enlarged  by  heating  until  it  will  slip  over  the  shaft,  in 
which  case  the  shrinkage  due 
to  cooling  causes  it  to  grip  the 
shaft. 

To  analyze  the  stresses  aris-     A         I  D9  I      Dl 
ing  from  shrinkage  and  forced 
fits,  let  Dl  denote  the  diameter 
of  the  hole  in  the  ring  or  disk,  j,      98 

and   D2  the   diameter   of   the 

shaft  (Fig.  98).  When  shrunk  or  forced  together,  Dl  must  increase 
slightly  and  Z>2  decrease  slightly;  that  is,  Dl  and  Z>2  must  of  necessity 
take  the  same  value  D.  Consequently,  the  circumference  of  the  hole 
changes  from  TT  Dl  to  TT  D,  and  hence  the  unit  deformation  81  of  a 
fiber  on  the  inner  surface  of  the  hole  is 


-  7rDl  _D  —  Dl 
i  i 


I) 


130  RESISTANCE  OF  MATERIALS 

Similarly,  the  unit  deformation  s2  of  a  fiber  on  the  surface  of  the 
shaft  is 


From  Hooke's  law,  —  —  E^  we  have,  therefore,  for  the  unit  stress 
s 

p1  on  the  inside  of  the  disk 


and  for  the  unit  stress  p2  on  the  surface  of  the  shaft 


—  s  = 


Adding  these  two  equations  to  eliminate  the  unknown  quantity  Z>, 
the  result  is 


where  K  denotes  the  allowance,  or  difference  in  diameter  of  shaft 
and  hole.  For  a  thick  disk  or  heavy  ring  this  allowance  K  may  be 
determined  from  the  nominal  diameter  D  of  the  shaft  by  means  of 
the  following  empirical  formulas :  * 

For  shrinkage  fits,        K  =  1&- 2.. 

9 

For  pressed  fits,  K  = 

For  driven  fits,  K  = 

For  thin  rings,  however,  the  allowance  given  by  these  formulas 
will  be  found  to  produce  stresses  in  the  ring  entirely  too  large  for 
safety.  In  deciding  on  the  allowance  for  any  given  class  of  work 
the  working  stresses  in  shaft  and  ring  may  first  be  assigned  and 
the  allowance  then  determined  from  the  formulas  given  below,  so 
that  the  actual  stresses  shall  not  exceed  these  values. 

*  S.  H.  Moore,  Trans.  Am.  Soc.  Mech.  Eng.,Vo\.  XXIV. 


SPHERES   AND   CYLINDERS  131 

From  Lame's  formulas  the  stresses  pl  and  p2  may  be  obtained  in 
terms  of  the  unit  pressure  between  the  surfaces  in  contact.  Thus, 
from  formula  (183),  the  stress  on  the  inside  of  the  hole  is 


00 


where  Z>g  denotes  the  outside  diameter  of  the  ring,  while,  by  substi- 
tuting r  =  a  and  b  =  0  in  the  equations  of  article  78,  the  stresses 
on  'the  outer  surface  of  the  shaft  are  found  to  be 

Ph  =  W>  Pr=  W' 

and  consequently 

P,  =  w- 
Eliminating  w  between  these  expressions  for  pl  and  p^  we  have 

(192)  H- 

fi 


Now,  to  simplify  the  solution,  let  the  coefficient  of  p2  be  denoted  by 
H;  that  is,  let 

A'+A* 

^=5F^' 

in  which  case 


Eliminating  p1  between  this  relation  and  the  above  expression  for 
the  allowance  /f,  we  have  finally 


(193) 


K 

P2  = 


+  — 


=  Hp 


In  applying  these  formulas  the  constant  If  is  first  computed  from 
the  given  dimensions  of  the  parts.  If  the  allowance  K  is  given, 
the  unit  stresses  pl  and  pz  in  ring  and  shaft  are  then  found  from  the 
above.  If  K  is  to  be  determined,  a  safe  value  for  the  stress  in  the 
ring  pl  is  assigned,  and  pz  is  calculated  from  the  second  equation. 
This  value  is  then  substituted  in  the  first  equation,  and  K  is 
calculated. 


RESISTANCE  'OF  MATERIALS 

APPLICATIONS 

216.  The  outside  diameter  of  a  pipe  is  4  in.,  and  thickness  of  wall  ^  in.  Find 
the  safe  internal  fluid  pressure  by  Clavarino's  formula  for  a  working  stress  in  the 
steel  of  10,000  lb./in.2 

Solution.  The  thickness  ratio  in  this  case  is  —  =  2  =  0.125  in.  Also,  Z>  =  4  in., 
d  -  3  in.,  p  =  10,000  lb./in.2,  and  consequently 

10(16-9)  _  2 


_  lb>/in> 


13  x  16  +  4  x  9 

217.  A  cast-iron  gear,  8  in.  external  diameter,  3  in.  wide,  and  l£in.  internal 
diameter,  is  to  be  forced  on  a  steel  shaft.  Find  the  stresses  developed,  the  pressure 
required  to  force  the  gear  on  the  shaft,  and  the  tangential  thrust  required  to  shear 
the  fit,  that  is,  to  produce  relative  motion  between  gear  and  shaft. 

Solution.   From  the  formula  K  =  —     —  *  the  allowance  is  found  to  be  .004  in., 

J.OUO 

making  the  diameter  of  the  shaft  J>2  =  1.754  in.  Also,  since  Dl  —  1.75  in.  and 
J>3  =  8  in.,  we  have  H=  1.1005.  Hence,  assuming  El  =  15,000,000  lb./in.2  and 
E2  =  30,000,000  lb./in.2,  we  have 

pl  =  23,550  lb./in.2,        p2  =  21,400  lb./in.2 

To  find  the  pressure  required  to  force  the  gear  on  the  shaft  it  is  first  necessary 
to  calculate  the  pressure  between  the  surfaces  in  contact.    From  the  relation 

p-  =w  this  amounts  to 

w  =  21,400  lb./in.2 

The  coefficient  of  friction  depends  on  the  nature  of  the  surfaces  in  contact.  As- 
suming it  to  be  fj.  =  .15  as  an  average  value,  and  with  a  nominal  area  of  contact 
of  TT  x  If  x  3  =  16.497  in.2,  the  total  pressure  P  required  is 

P  =  16.497  x  21,400  x  .15  =  52,955  Ib.  =  26.5  tons. 
To  find  the  torsional  resistance  of  the  fit,  we  have,  as  above, 

Bearing  area  =  16.497  in.2,         Unit  pressure  =  21,400  lb./in.2, 

fjL  =  .15,  radius  of  shaft  =  .875  in. 
Hence  the  torsional  resistance  is 

Mt  =  16.497  x  21,400  x  .15  x  .875  =  46,336  in.-lb. 

Consequently  the  tangential  thrust  on  the  teeth  of  the  gear  necessary  to  shear  the 
4£3iiL  =  n,584  Ib.  =  5.8  tons. 

218.  The  outside  diameter  of  a  steel  pipe  is  5^  in.,  thickness  of  wall  1  in.,  and 
internal  fluid  pressure  1500  lb./in.2    Find  by  Clavarino's  formula  the  maximum 
fiber  stress  in  the  wall  of  the  pipe. 

219.  The  outside  diameter  of  a  steel  pipe  is  8  in.,  the  internal  fluid  pressure  is 
2000  lb./in.2,  and  the  allowable  stress  in  the  steel  is  15,000  lb./in.2  Find  the  required 
thickness  of  pipe  wall. 

220.  Solve  problem  216  by  the  other  four  formulas  listed  in  article  79  and  com- 
pare the  results. 


SPHERES  AND   CYLINDERS 


133 


FIG.  99 


221.  Find  the  thickness  necessary  to  give  to  a  steel  locomotive  cylinder  of  22  in. 
internal  diameter  if  it  is  required  to  withstand  a  maximum  steam  pressure  of 
1501b./in.2  with  a  factor  of  safety  of  10,  using  both  Lamp's  and  Clavarino's  formulas. 

222.  In  a  four-cycle  gas  engine  the  cylinder  is  of  steel 
with  an  internal  diameter  of  6  in.,  and  the  initial  internal 
pressure  is  200  lb./in.2  absolute.    With  a  factor  of  safety 
of  15,  how  thick  should  the  walls  of  the  cylinder  be  made, 
according  to  Lamp's  formula  ? 

223.  The  steel  cylinder   of  a  hydraulic   press  has  an 
internal  diameter  of  5  in.  and  an  external  diameter  of  7  in. 
With  a  factor  of  safety  of  3,  how  great  an  internal  pressure 
can  the  cylinder  withstand,  according  to  Lamp's  formula  ? 

224.  In  a  fire-tube  boiler  the  tubes  are  of  drawn  steel, 
2  in.  internal  diameter  and  |  in.  thick.   What  is  the  factor 
of  safety  for  a  working  gauge  pressure  of  200  lb./in.2  ? 

225.  How  great  is  the  stress  in  a  copper  sphere  2  ft.  in 
diameter  and  .25  in.  thick  under  an  internal  pressure  of 
175  lb./in.2  ? 

226.  A  cast-iron  water  pipe  is  24  in.  in  diameter  and 
2  in.  thick.    What  is  the  greatest  internal  pressure  which  it 
can  withstand,  according  to  the  formula  for  thin  cylinders  ? 

227.  A  wrought-iron  cylinder  is  8  in.  in  external  diame- 
ter and  1£  in.  thick.    How  great  an  external  pressure  can 
it  withstand  ? 

228.  An  elevated  water  tank  is  cylindrical  in  form,  with 

a  hemispherical  bottom  (Fig.  99).   The  diameter  of  the  tank  is  20  ft.  and  its  height 
52  ft.  (exclusive  of  the  bottom).    If  the  tank  is  to  be  built  of  wrought  iron  and  the 
factor  of  safety  is  taken  to  be  6,  what  should  be  the  thickness  of  the  bottom  plates 
and  also  of  those  in  the  body  of  the  tank  near  its  bottom  ? 

NOTE.  Formulas  (165)  and  (166)  give  the  required  thick- 
ness of  the  plates,  provided  the  tank  is  without  joints.  The 
bearing  power  of  the  rivets  at  the  joints,  however,  is,  in  general, 
the  consideration  which  determines  the  thickness  of  the  plates 
(article  90). 

229.  A  marine  boiler  shell  is  16  ft.  long,  8  ft.  in  diam- 
eter, and  1  in.  thick.    What  is  the  stress  in  the  shell  for  a 
working  gauge  pressure  of  160  lb./in.2  ? 

230.  The  air  chamber  of  a  pump  is  made  of  cast  iron 
of  the  form  shown  in  Fig.  100.    If  the  diameter  of  the  air 
chamber  is  10  in.  and  its  height  24  in.,  how  thick  must  the 
walls  of  the  air  chamber  be  made  in  order  to  stand  a  pres- 
sure of  500  lb./in.2,  with  a  factor  of  safety  of  4  ? 

231.  The  end  plates  of  a  boiler  shell  are  curved  out  to 
a  radius  of  5  ft.  If  the  plates  are  f  in.  thick,  find  the  tensile 
stress  due  to  a  steam  pressure  of  175  lb./in.2 

232.  If  the  thickness  of  the  end  plates  in  problem  231 

is  changed  to  ^  in.,  the  steam  pressure  being  the  same,  to  what  radius  should 
they  be  curved  in  order  that  the  tensile  stress  in  them  shall  remain  the  same  ? 


FIG. 100 


134 


RESISTANCE  OF  MATERIALS 


233.  The  cylinder  of  an  hydraulic  press  is  12  in.  inside  cliam.    How  thick  must 
it  be  in  order  to  stand  a  pressure  of   1500  lb./in.2  if  it  is  made  of  cast  steel  and 
the  factor  of  safety  is  10  ? 

234.  A  high-pressure  cast-iron  water  main  is  4  in.  inside  diameter  and  carries 
a  pressure  of  800  lb./in.2   Find  its  thickness  for  a  factor  of  safety  of  15. 

235.  The  water  chamber  of  a  tire  engine  has  a  spherical  top  18  in.  in  diameter 
and  carries  a  pressure  of  250  lb./in.2    It  is  made  of  No.  7  B.  and  S.  gauge  copper, 
which  is  reduced  in  manufacture  to  a  thickness  of  about  .1  in.   Determine  the  fac- 
tor of  safety. 

236.  A  cast-iron  ring  3  in.  thick  and  8  in.  wide  is  forced  onto  a  steel  shaft  10  in. 
in  diameter.  Find  the  stresses  in  ring  and  shaft,  the  pressure  required  to  force  the 
ring  onto  the  shaft,  and  the  torsional  resistance  of  the  fit. 

NOTE.  Since  the  ring  in  this  case  is  relatively  thin,  assume  an  allowance  of  about  half 
the  amount  given  by  Moore's  formula.  Then,  having  given  7)2  —  10  in.,  />3  =  16  in.,  and 
having  computed  the  allowance  A",  we  have  also  Dl  —  D2  —  K,  and,  inserting  these  values 
in  the  formulas  of  article  82,  the  required  quantities  may  be  found,  as  explained  in 
problem  217. 

237.  The  following  data  are  taken  from  Stewart's  experiments  on  the  collapse 
of  thin  tubes  under  external  pressure,  the  tubes  used  for  experiment  being  lap- 
welded  steel  boiler  flues.     Compute  the   collapsing  pressure   from  the  rational 
formula  for  thin  tubes,  given  in  article  81,  for  both  the  average  thickness  and  least 
thickness,  and  note  that  these  two  results  lie  on  opposite  sides  of  the  value  obtained 
directly  by  experiment. 


OUTSIDE  DIAMETER  IN  INCHES 

THICKNESS  h  IN  INCHES 

ACTUAL 

At  place  of  collapse 

At  place  of  collapse 

ING 

Average 

Greatest  =  D 

Least  =  <1 

Greatest 

Least 

Ib./in.a 

8.  (504 

8.610 

8.580 

0.219 

0.230 

0.210 

870 

8.664 

8.670 

8.625 

0.226 

0.227 

0.204 

840 

8.665 

8.670 

8.660 

0.212 

0.240 

0.211 

880 

8.653 

8.665 

8.590 

0.208 

0.220 

0.203 

970 

8.688 

8.715 

8.605 

0.274 

0.280 

0.266 

1430 

8.664 

8.695 

8.635 

0.258 

0.261 

0.248 

1320 

8.645 

8.665 

8.635 

0.263 

0.268 

0.259 

1590 

8.674 

8.675 

8.675 

0.273 

0.282 

0.270 

2030 

8.638 

8.645 

8.615 

0.289 

0.298 

0.280 

2200 

10.055 

10.180 

9.950 

0.157 

0.182 

0.150 

210 

238.  The  following  data  are  taken  from  Stewart's  experiments  on  the  collapse 
of  thick  tubes  under  external  pressure.  The  ultimate  compressive  strength  of  the 
material  was  not  given  by  the  experimenter,  but  from  the  other  elastic  properties 
given  it  is  here  assumed  to  be  uc  =  38,500  lb./in.2  Compute  the  collapsing  pres- 
sure from  the  rational  formula  for  thick  tubes,  given  in  article  81,  for  both  average 
and  least  thickness,  and  compare  these  results  with  the  actual  collapsing  pressure 
obtained  by  experiment. 


SPHERES  AND   CYLINDERS 


135 


OUTSIDE  DIAMETER  IN  INCHES 

THICKNESS  h  IN  INCHES 

ACTUAL 

At  place  of  collapse 

At  place  of  collapse 

ING 

Greatest  =D 

Least  =  (I 

Greatest 

Least 

lb./in.2 

4.010 

4.020 

3.980 

0.173 

0.203 

0.140 

2050 

4.014 

4.050 

3.990 

0.178 

0.277 

0.158 

2225 

4.012 

4.050 

3.960 

0.173 

0.200 

0.170 

2425 

4.018 

4.050 

4.010 

0.184 

0.192 

0.165 

2540 

2.997 

3.010 

2.980 

0.147 

0.151 

0.138 

3350 

2.987 

3.010 

2.970 

0.139 

0.139 

0.125 

2575 

2.990 

3.010 

2.970 

0.190 

0.218 

0.166 

4200 

2.996 

3.020 

2.980 

0.191 

0.216 

0.176 

4200 

2.997 

3.020 

2.960 

0.190 

0.215 

0.161 

4175 

3.000 

3.020 

2.960 

0.182 

0.192 

0.165 

3700 

239.  What  is  the  maximum  external  pressure  which  a  cast-iron  pipe  18  in.  in 
diameter  and  \  in.  thick  can  stand  without  crushing  ? 

240.  Solve  problem  226  by  Birnie's  and  Barlow's  formulas. 


SECTION  XI 

FLAT  PLATES 

83.  Theory  of  flat  plates.    The  analysis  of  stress  in  flat  plates  is 
at  present  the  most  unsatisfactory  part  of  the  strength  of  materials. 
Although  flat  plates  are  of  frequent  occurrence  in  engineering  con- 
structions (as,  for  example,  in  manhole  covers,  cylinder  ends,  floor 
panels,  etc.),  no  general  theory  of  such  plates  has  as  yet  been  given. 
Each  form  of  plate  is  treated  by  a  special  method,  which  in  most 
cases  is  based  upon  an  arbitrary  assumption  either  as  to  the  danger- 
ous section  or  as  to  the  reactions  of  the  supports,  and  therefore 
leads  to  questionable  results. 

Although  the  present  theory  of  flat  plates  is  plainly  inadequate, 
it  is  nevertheless  of  value  in  pointing  out  the  conditions  to  which 
such  plates  are  subject,  and  in  furnishing  a  rational  basis  for  the 
estimation  of  their  strength.  The  formulas  derived  in  the  following 
paragraphs,  if  used  in  this  way,  with  a  clear  understanding  of  their 
approximate  nature,  will  be  found  to  be  invaluable  in  designing, 
or  in  determining  the  strength  of  flat  plates. 

The  following  has  come  to  be  the  standard  method  of  treatment 
and  is  chiefly  due  to  Bach.* 

84.  Maximum  stress  in  homogeneous  circular  plate  under  uni- 
form load.    Consider  a  flat,  circular  plate  of  homogeneous  ma- 
terial, which  bears  a  uniform  load  of  amount  w  per  unit  of  area, 
and  suppose  that  the  edge  of  the  plate  rests  freely  on  a  circular 
rim  slightly  smaller  than  the  plate,  every  point  of  the  rim  being 
maintained  at  the  same  level.    The  strain  in  this  case  is  greater 
than  it  would  be  if  the  plate  was  fixed  at  the  edges,  and  conse- 
quently the  formula  deduced  will  give  the  maximum  stress  in 
all  cases. 

*  For  an  approximate  method  of  solution  see  article  by  S.  E.  Slocum  entitled  "  The 
Strength  of  Flat  Plates,  with  an  Application  to  Concrete-Steel  Floor  Panels,"  Engineer- 
ing News,  July  7,  1904. 

136 


FLAT  PLATES 


137 


Now  suppose  a  diametral  section  of  the  plate  taken,  and  regard 
either  half  of  the  plate  as  a  cantilever  (Fig.  101).    Then  if  r  is  the 

2 

radius  of  the  plate,  the  total  load  on  this  semicircle  is  -  —  w,  and 

2 

its  resultant  is  applied  at  the  center  of  gravity  of  the  semicircle, 
which  is  at  a  distance  of  - —  from  AB.    The  moment  of  this  result- 

O  7T  2  A  f)      3 

ant  about  the  support  AB  is  therefore  -—w  •  — — ,  or  -    —    Simi- 

2  67T  O 

larly,  the  resultant  of  the  supporting  forces  at  the  edge  of  the 

plate  is  of  amount  -^-  •  w  and  is  applied  at  the  center  of  gravity  of 

2r 
the  semi-circumference,  which  is  at  a  distance  of  —  from  AB.    The 

moment  of  this  resultant  about  AB  is  therefore 


irr  w     z  r 


,  or  r*w.    Hence  the  total  external 


2  7T 

moment  M  at  the  support  is 

2  r*w      r8w 
M^Sw-  —    -—. 

Now  assume  that  the  stress  at  any  point  of 
the  plate  is  independent  of  the  distance  of 
this  point  from  the  center.    Under  this  arbi- 
trary assumption  the  stress  in  the  plate  is  given  by  the  fundamental 
formula  in  the  theory  of  beams,  namely, 

Me 


FIG. 101 


If  the  thickness  of  the  plate  is  denoted  by  A,  then,  since  the  breadth 
of  the  section  is  b  —  2  r, 

bh*      rW  h 


Consequently, 


r\v     h 
Me      ~3~"  2 


p  =  —  = 


rh 


whence 
(194) 


138  RESISTANCE  OF  MATERIALS 

Foppl  has  shown  that  the  arbitrary  assumption  made  in  deriving 
this  formula  can  be  avoided,  and  the  same  result  obtained,  by  a  more 
rigorous  analysis  than  the  preceding,  and  Bach  has  verified  the 
formula  experimentally.  Formula  (194)  is  therefore  well  established 
both  theoretically  and  practically. 

85.  Maximum  stress  in  homogeneous  circular  plate  under  con- 
centrated load.  Consider  a  flat,  circular  plate  of  homogeneous  mate- 
rial, and  suppose  that  it  bears  a  single  concentrated  load  P  which  is 
distributed  over  a  small  circle  of  radius  rQ  concentric  with  the  plate. 
Taking  a  section  through  the  center  of  the  plate  and  regarding  either 
half  as  a  cantilever,  as  in  the  preceding  article,  the  total  rim  pres- 

~p  2  T 

sure  is  —  »  and  it  is  applied  at  a  distance  of  —  from  the  center.    The 

P   * 
total  load  on  the  semicircle  of  radius  rQ  is  —  ,  and  it  is  applied  at  a  dis- 

4r 

tance  of  —  -  from  the  section.   Therefore  the  total  external  moment  M 

6  7T 

at  the  section  is         _  Pr      2  Pr0  _  Pr  L      2  r0 

7T  3  7T  7T    \  3r 

Assuming  that  the  stress  is  uniformly  distributed  throughout  the 
plate,  the  stress  due  to  the  external  moment  M  is  given  by  the 
formula  •Me 

p=-. 

If  the  thickness  of  the  plate  is  denoted  by  A,  then 

rW  h 

/=_     and     e  =  -. 

Therefore  p    /        %    \  -L 

_[  1  _      °  ) 
Me       TT  \        3r/2 


whence 
(195) 


3 


If  rQ  =  0,  that  is  to  say,  if  the  load  is  assumed  to  be  concentrated 
at  a  single  point  at  the  center  of  the  plate,  formula  (195)  becomes 


(196) 


FLAT  PLATES 


139 


If  the  load  is  uniformly  distributed  over  the  entire  plate,  then 
rQ  =  r  and  P  =  TTT^W,  where  w  is  the  load  per  unit  of  area.  In  this 
case  formula  (195)  becomes 


p  = 


2\  IT 

-s  rwd 


which  agrees  with  the  result  of  the  preceding  article. 

86.  Dangerous  section  of  elliptical  plate.  Consider  a  homogeneous 
elliptical  plate  of  semi-axes  a  and  b  and  thickness  7i,  and  suppose 
that  an  axial  cross  is  cut  out  of  the  plate,  composed  of  two  strips 
AB  and  CD,  each  of  unit  width,  in- 
tersecting in  the  center  of  the  plate, 
as  shown  in  Fig.  102. 

Now  suppose  that  a  single  concen- 
trated load  acts  at  the  intersection 
of  the  cross  and  is  distributed  to  the 
support  in  such  a  way  that  the  two 
beams  AB  and  CD  each  deflect  the 
same  amount  at  the  center.  Since 
AB  is  of  length  2  a,  from  article  40,  equation  (54),  the  deflection 

P  (  2  aV 

at  the  center  of  AB  is  Dl  —  — ^ — ~~-  From  symmetry,  the  reac- 
tions at  A  and  B  are  equal.  Therefore,  if  each  of  these  reactions  is 
denoted  by  R^  2  R^  =  P  and,  consequently, 


_ 

~ 


Similarly,  if  R^  denotes  the  equal  reactions  at  C  and  D,  the  deflec- 


tion DZ  of  CD  at  its  center  is 


If  the  plate  remains  intact,  the  two  strips  AB  and  CD  must  deflect 
the  same  amount  at  the  center.    Therefore  Dl  =  Z>2,  and  hence 


(197) 


-* 


140  RESISTANCE  OF  MATERIALS 

For  the  beam  AB  of  length  2  a  the  maximum  external  moment  is 

H^t.  Also,  since  AB  is  assumed  to  be  of  unit  width,  1=  — —  and  e  =  —  • 
Hence  the  maximum  stress  p'  in  AB  is 


Similarly,  the  maximum  stress  p"  in  CD  is 

I 

Consequently, 


p" 


or,  since  from  equation  (197)  —  =  —  • 

Rn         CL 


/_ 


By  hypothesis  a  >  b.    Therefore  p"  >  p'  \  that  is  to  say,  the  maxi- 
mum stress  occurs  in  the  strip  CD  (that  is,  in  the  direction  of  the 
shorter  axis  of  the  ellipse).   In  an  elliptical  plate,  therefore,  rupture 
may  be  expected  to  occur  along  a  line  parallel  to  the  major  axis  — 
a  result  which  has  been  confirmed  by  experiment. 

87.  Maximum  stress  in  homogeneous  elliptical  plate  under  uniform 
load.  The  method  of  finding  the  maximum  stress  in  an  elliptical 
plate  is  to  consider  the  two  limiting  forms  of  an  ellipse,  namely, 
a  circle  and  a  strip  of  infinite  length,  and  express  a  continuous 
relation  between  the  stresses  for  these  two  limiting  forms.  The 
method  is  therefore  similar  to  that  used  in  Article  54  in  obtaining 
the  modified  form  of  Euler's  column  formula. 

Consider  first  an  indefinitely  long  strip  with  parallel  sides, 
supported  at  the  edges  and  bearing  a  uniform  load  of  amount 
w  per  unit  of  area.  Let  the  width  of  the  strip  be  denoted  by  25 
and  its  thickness  by  h.  Then,  if  this  strip  is  cut  into  cross  strips 
of  unit  width,  each  of  these  cross  strips  can  be  regarded  as  an 
independent  beam,  the  load  on  one  of  these  unit  cross  strips 

being  2bw  and  the  maximum  moment  at  the  center  being  - — - — 


FLAT  PLATES  141 

Consequently,  the  maximum  stress  in  the  cross  strips,  and  therefore 
in  the  original  strip,  is  4 

(198) 


In  the  preceding  article  it  was  shown  that  the  maximum  stress  in 
an  elliptical  plate  occurs  in  the  direction  of  the  minor  axis.  There- 
fore equation  (198)  gives  the  limiting  value  which  the  stress  in  an 
elliptical  plate  approaches  as  the  ellipse  becomes  more  and  more 
elongated. 

For  a  circular  plate  of  radius  b  and  thickness  h  the  maximum 
stress  was  found  to  be 


Comparing  equations  (198)  and  (199),  it  is  evident  that  the  maxi- 
mum stress  in  an  elliptical  plate  is  given,  in  general,  by  the  formula 


where  k  is  a  constant  which  lies  between  1  and  3.    Thus,  for  -  =  1 

,  a 

(that  is,  for  a  circle)  k=  1  ;  whereas,  if  -  =  0  (that  is,  for  an  infinitely 

long  ellipse),  k  =  3.  The  constant  k  may  therefore  be  assumed  to 
have  the  value  7 

*=8-2-, 

a 

which  reduces  to  the  values  1  and  3  for  the  limiting  cases,  and  in 
other  cases  has  an  intermediate  value  depending  on  the  form  of  the 
plate.  Consequently, 

(3a- 
(200) 

which  is  the  required  formula  for  the  maximum  stress  p  in  a  homo- 
geneous elliptical  plate  of  thickness  h  and  semi-axes  a  and  b. 

88.  Maximum  stress  in  homogeneous  square  plate  under  uniform 
load.  In  investigating  the  strength  of  square  plates  the  method  of 
taking  a  section  through  the  center  of  the  plate  and  regarding  the 


142 


RESISTANCE  OF  MATERIALS 


portion  of  the  plate  on  one  side  of  this  section  as  a  cantilever  is 
used,  but  experiment  is  relied  upon  to  determine  the  position  of  the 
dangerous  section.  From  numerous  experiments  on  flat  plates  Bach 
has  found  that  homogeneous  square  plates  under  uniform  load 
always  break  along  a  diagonal.* 

Consider  a  homogeneous  square  plate  of 
thickness  h  and  side  2  a,  which  bears  a 
uniform  load  w  per  unit  of  area.  Suppose 
that  a  diagonal  section  of  this  plate  is 
taken,  and  consider  either  half  as  a  canti- 
lever, as  shown  in  Fig.  103.  Then  the  total 
load  on  the  plate  is  4wa2,  and  the  reac- 
tion of  the  support  under  each  edge  is  wa2. 
If  d  denotes  the  length  of  the  diagonal  AC,  the  resultant  pres- 
sure on  each  edge  of  the  plate  is  applied  at  a  distance  —  from  AC, 

7 

and  therefore  the  moment  of  these  resultants  about  AC  is  2  (wa2)  -, 

2  ?  4 

or  — —    The  total  load  on  the  triangle  ABC  is  2  wa2,  and  its  result- 

Li 

ant  is  applied  at  the  center  of  gravity  of  the  triangle,  which  is  at  a 

distance  of  -  from  AC.    Therefore  the  moment  of  the  load  about 
b   ,  2 , 

AC  is  (2  wa2)-,  or  -    —    Therefore  the  total  external  moment  M 
b  o 


FIG. 103 


at  the  section  A  C  is 


wa2d      wa2d     wa2d 


236 
Hence  the  maximum  stress  in  the  plate  is 

wa2d    h 
2 


from  which 
(201) 


The  maximum  stress  in  a  square  plate  of  side  2  a  is  therefore  the 
same  as  in  a  circular  plate  of  diameter  2  a. 

*  Bach,  Elasticitdt  und  Festigkeitslehre,  3d.  ed.,  p.  561. 


FLAT  PLATES 


143 


89.  Maximum  stress  in  homogeneous  rectangular  plate  under 
uniform  load.  In  the  case  of  rectangular  plates  experiment  does 
not  indicate  so  clearly  the  position  of  the  dangerous  section  as 'it 
does  for  square  plates.  It  will  be  assumed  in  what  follows,  how- 
ever, that  the  maximum  stress  occurs  along  a  diagonal  of  the  rec- 
tangle. This  assumption  is  at  least  approximately  correct  if  the  length 
of  the  rectangle  does  not  exceed  two  or  three  times  its  breadth. 

Let  the  sides  of  the  rectangle  be  denoted  by  2  a  and  2  b,  and  the 
thickness  of  the  plate  by  h  (Fig.  104).  Also  let  d  denote  the 
length  of  the  diagonal  AC,  and  c  oa 

the  altitude  of  the  triangle  ABC. 
Now  suppose  that  a  diagonal  sec- 
tion AC  of  the  plate  is  taken,  and 
consider  the  half  plate  ABC  as  a 
cantilever,  as  shown  in  Fig.  104. 
If  w  denotes  the  unit  load,  the 
total  load  on  the  plate  is  4  abw,  and 
consequently  the  resultant  of  the 
reactions  of  the  supports  along 
AB  and  BC  is  of  amount  2  abw  and 

s* 

is  applied  at  a  distance  -  from  A  C. 

Therefore  the  moment  of  the  sup- 
porting force  about  AC  is  abwc. 
Also,  the  total  load  on  the  triangle  ABC  is  2  abw,  and  it  is  applied 

at  the  center  of  gravity  of  the  triangle,  which  is  at  a  distance  of  — 

o 

from  AC.    Consequently,  the  total  moment  of  the  load  about  AC  is 
— .    Therefore  the  total  external  moment  M  at  the  section  AC  is 


FIG.  104 


M  =  abwc  — 


2  abwc      abwc 


3  3 

and  the  maximum  stress  in  the  plate  is 

abwc    h 

_  Me      ~3~  '  2      2  wabc 
I 


dtf 
12 


dtt 


144  RESISTANCE  OF  MATERIALS 

or,  since  cd  =  4  ab, 

(202)  p  =  w£-, 

which  gives  the  required  maximum  stress. 

For  a  square  plate  a  =  b  and  c  =  a  V2,  and  formula  (202)  reduces 
to  formula  (201)  for  square  plates,  obtained  in  the  preceding  article. 

APPLICATIONS 

241.  The  cylinder  of  a  locomotive  is  20  in.  internal  diameter.    What  must  be 
the  thickness  of  the  steel  end  plate  if  it  is  required  to  withstand  a  pressure  of 
160  lb./in.2  with  a  factor  of  safety  of  6  ? 

242.  A  circular  cast-iron  valve  gate  ^  in.  thick  closes  an  opening  6  in.  in  diam- 
eter.   If  the  pressure  against  the  gate  is  due  to  a  water  head  of  150  ft.,  what  is  the 
maximum  stress  in  the  gate  ? 

243.  Show  that  the  maximum  concentrated  load  which  can  be  borne  by  a 
circular  plate  is  independent  of  the  radius  of  the  plate. 

244.  A  cast-iron  manhole  cover  1  in.  thick  is  elliptical  in  form  and  covers  an  ellip- 
tical opening  3  ft.  long  and  18  in.  wide.   How  great  a  uniform  pressure  will  it  stand  ? 

245.  What  must  be  the  thickness  of  a  wrought-iron  plate  covering  an  opening 
4  ft.  square  in  order  to  carry  a  load  of  200  lb./ft.2  with  a  factor  of  safety  of  5  ? 

246.  A  wrought-iron  trap  door  is  5  ft.  long,  3  ft.  wide,  and  |  in.  thick.    How 
great  a  uniform  load  will  it  bear  ? 

247.  The  steel  diaphragm  separating  two  expansion  chambers  of  a  steam  turbine 
is  subjected  to  a  pressure  of  150  lb./in.2  on  one  side  and  80  lb./in.2  on  the  other. 
Find  the  required  thickness  for  a  factor  of  safety  of  10. 

248.  The  cylinder  of  a  hydraulic  press  is  made  of  cast  steel,  10  in.  inside  diam- 
eter, with  a  flat  end  of  the  same  thickness  as  the  walls  of  the  cylinder.    Find  the 
required  thickness  for  a  factor  of  safety  of  20.    Also  find  how  much  larger  the 
factor  of  safety  would  be  if  the  end  was  made  hemispherical.  Assume  w  =  1200  lb./in.2 

249.  The  cylinder  of  a  steam  engine  is  16  in.  inside  diameter  and  carries  a  steam 
pressure  of  125  lb./in.2   If  the  cylinder  head  is  mild  steel,  find  its  thickness  for  a 
factor  of  safety  of  10. 

250.  A  cast-iron  valve  gate  10  in.  in  diameter  is  under  a  pressure  head  of  200  ft. 
Find  its  thickness  for  a  factor  of  safety  of  15. 

251.  A  cast-iron  elliptical  manhole  cover  is  18  in.  x  24  in.  in  size  and  is  designed 
to  carry  a  concentrated  load  of  1000  Ib.    If  the  cover  is  ribbed,  how  thick  must  it 
be  for  a  factor  of  safety  of  20,  assuming  that  the  ribs  double  its  strength  ? 

252.  Thurston's  rule  for  the  thickness  of  cylinder  heads  for  steam  engines  is 

h  =  .00035  wD, 

where  h  =  thickness  of  head  in  inches, 

D  =  inside  diameter  of  cylinder  in  inches, 
w  =  pressure  in  lb./in.2 

Compare  this  formula  with  Bach's,  assuming  the  material  to  be  wrought  iron  and 
using  the  data  of  problem  249. 


FLAT  PLATES 


145 


253.  Show  that  Thurston's  rule  for  thickness  of  cylinder  head,  given  in  problem 
252,  makes  thickness  of  head  =  1^  times  thickness  of  walls. 

254.  Nichols's  rule  for  the  proper  thickness  of  unbraced  flat  wrought-iron  boiler 
heads  is 


h  = 


Aw 
IQp 


where 


h  =  thickness  of  head  in  inches, 
A  =  area  of  head  in  square  inches, 
w  =  pressure  per  square  inch, 

50,000      ultimate  strength  in  tension 

p  =  working  stress  =  —  — 

10  factor  of  safety 

Compare  this  empirical  rule  with  Bach's  formula,  using  the  data  of  problem  249 
and  assuming  the  material  to  be  wrought  iron. 

255.  Nichols's  rule  for  the  collapsing  pressure  of  unbraced  flat  wrought-iron 
boiler  heads  is 

1U  llUt 

— • 

where    w  =  collapsing  pressure  in  lb./in.2,  h  =  thickness  of  head  in  inches, 

ut  =  ultimate  tensile  strength  in  lb./in.2,  A  =  area  of  head  in  square  inches. 
Show  that  Nichols's  two  formulas  are  identical  and  that  therefore  they  cannot  be 
rational. 

256.  The  following  data  are  taken  from  Nichols's  experiments  on  flat  wrought- 
iron  circular  plates. 


DIAMETER 

THICKNESS 

ACTUAL  BURSTING 

IN  INCITES 

IN  INCHES 

PRESSURE  LB./IN.S 

34.5 

T9(T 

280 

34.5 

1 

200 

28.5 

3 

5 

300 

26.5 

3 

370 

Using  these  data,  compare  Bach's  and  Grashof's  rational  formulas  with  Nichols's 
and  Thurston's  empirical  formulas,  as  given  below  : 

Circular  plate,  supported  at  edge  and  uniformly  loaded. 


Bach, 

Grashof, 


fw 

h  =  r\-  =  .5D 
\ 


=  .4564DX/-, 
\p 


Nichols, 
Thurston, 
where 


10  p  p 

h  =  .00035  w D, 

h  =  thickness  of  head  in  inches,       D  =  diameter  of  head  in  inches  =  2  r, 
w  =  pressure  in  lb./in.2,  p  =  working  stress  in  lb./in.2, 

A  =  area  of  head  in  square  inches  =  —  —  • 


Note  that  the  Nichols  and  Thurston  formulas  apply  only  to  wrought  iron. 


SECTION  XII 

RIVETED  JOINTS  AND  CONNECTIONS 

90.  Efficiency  of  riveted  joint.  In  structural  work  such  as  plate 
girders,  trusses,  etc.,  and  also  in  steam  boilers,  standpipes,  and 
similar  constructions,  the  connections  between  the  various  members 
are  made  by  riveting  the  parts  together.  Since  the  holes  for  the 
rivets  weaken  the  members  so  joined,  the  strength  of  the  structure 
is  determined  by  the  strength  of  the  joint. 

Failure  of  a  riveted  joint  may  occur  in  various  ways ;  namely,  by 
shearing  across  the  rivet,  by  crushing  the  rivet,  by  crushing  the 
plate  in  front  of  the  rivet,  by  shearing  the  plate  (that  is,  pulling 
out  the  rivets),  or  by  tearing  the  plate  along  the  line  of  rivet  holes. 
Experience  has  shown,  however,  that  failure  usually  occurs  either 
by  shearing  across  the  rivet  or  by  tearing  the  plate  along  the  line 
of  rivet  holes. 

The  strength  of  any  given  type  of  riveted  joint  is  expressed  by 
what  is  called  its  efficiency,  denned  as 

strength  of  joint 
Efficiency  of  riveted  joint  = 


strength  of  unriveted  member 

Thus,  in  Fig.  105,  if  d  denotes  the  diameter  of  a  rivet  and  c  the 
distance  between  rivet  holes,  or  pitch  of  the  rivets,  as  it  is  called, 
the  efficiency,  e,  of  the  joint  against  tearing  of  the  plate  along  the 
line  of  rivet  holes  is 

c-d 

e  = 

c 

To  determine  the  efficiency  of  the  joint  against  shearing  across  the 
rivets,  let  q  denote  the  ultimate  shearing  strength  of  the  rivet  and  p 
the  ultimate  tensile  strength  of  the  plate.  Then,  for  a  single-riveted 
lap  joint  (Fig.  105),  if  h  denotes  the  thickness  of  the  plate,  the 
area  corresponding  to  one  rivet  is  lie,  and  the  area  in  shear  for 

146 


RIVETED  JOINTS  AND  CONNECTIONS 


147 


each  rivet  is  -  —  •    Consequently  the  efficiency  of  this  type  of  joint 
against  rivet  shearing  is  72 


e  = 


4  chp 


For  an  economical  design  these  two  efficiencies  should  be  equal.  For 
practical  reasons,  however,  it  is  not  generally  possible  to  make  these 


1         ,1 

>c—  at*- 


SINGLE-RIVETED  LAP  JOINT 
EFFICIENCY  50-60  PER  CENT 


SINGLE-RIVETED  BI:TT  JOINT 
EFFICIENCY  76-78  PER  CENT 


©  ©  © 
©  ©  © 


DOUBLE-RIVETED  LAP  JOINT  DOUBLE-RIVETED  BUTT  JOINT 

EFFICIENCY  70-72  PER  CENT  EFFICIENCY  82-83  PER  CENT 

FIG. 105 

exactly  equal,  and  in  this  case  the  smaller  of  the  two  determines 
the  strength  of  the  joint. 

For  a  double-riveted  lap  joint  the  efficiency  against  tearing  of 

the  plate  is 

G—  d 


148  RESISTANCE  OF  MATERIALS 

as  above,  but  since  in  this  case  there  are  two  rivets  for  each  strip 
of  length  <?,  the  efficiency  against  rivet  shear  is 


Similarly,  for  a  single-riveted  butt  joint  with  two  cover  plates 
the  efficiency  of  the  joint  against  tearing  of  the  plate  is 

c  —  d 

e  =  -  , 
c 

and  against  rivet  shear  is 

=  ird*q 
~  2chp' 

For  a  double-riveted  butt  joint  with  two  cover  plates  the  efficiency 
against  tearing  of  the  plate  is 

_  c—  d 

and  against  rivet  shear  is 


chp 

The  average  efficiencies  of  various  types  of  riveted  joints  as  used 
in  steam  boilers  are  given  in  Fig.  105. 

91.  Boiler  shells.  In  designing  steam-boiler  shells  it  is  customary 
in  this  country  to  determine  first  the  thickness  of  shell  plates,  by 
the  following  rule  : 

To  find  the  thickness  of  shell  plates,  multiply  the  maximum  steam 
pressure  to  be  carried  (safe  working  pressure  in  lb./  in.2)  by  half  the 
diameter  of  the  boiler  in  inches.  This  gives  the  hoop  stress  in  the 
shell  per  unit  of  length.  Divide  this  result  by  the  safe  working  stress 
(working  stress  =  ultimate  strength,  usually  about  60,000  lb./in.2, 
divided  by  the  factor  of  safety,  say  4  or  5),  and  divide  the  quotient 
by  the  average  efficiency  of  the  style  of  joint  to  be  used,  expressed 
as  a  decimal.  The  result  will  be  the  thickness  of  the  shell  plates 
expressed  in  decimal  fractions  of  an  inch. 

Having  determined  the  thickness  of  shell  plates  by  this  method, 
the  diameter  of  the  rivets  is  next  found  from  the  empirical  formula 


KIVETED  JOINTS  AND   CONNECTIONS  149 

where  k=  1.5  for  lap  joints  and  k=  1.3  for  butt  joints  with  two 
cover  plates. 

The  pitch  of  the  rivets  is  next  determined  by  equating  the  strength 
of  the  plate  along  a  section  through  the  rivet  holes  to  the  strength 
of  the  rivets  in  shear  and  solving  the  resulting  equation  for  c. 

To  illustrate  the  application  of  these  rules,  let  it  be  required  to 
design  a  boiler  shell  48  in.  in  diameter  to  carry  a  steam  pressure  of 
125  lb./in.2  with  a  double-riveted,  double-strapped  butt  joint. 

By  the  above  rule  for  thickness  of  shell  plates  we  have 

125  x  -4- 


5 

The  diameter  of  rivets  is  then 

d  =  1.3>/T:=.73,  say  fin. 

To  determine  the  pitch  of  the  rivets,  the  strength  of  the  plate  for 
a  section  of  width  c  on  a  line  through  the  rivet  holes  is 

(c-£)hp  =  (c-  |)  ^  x  60,000, 
and  the  strength  of  the  rivets  in  shear  for  a  strip  of  this  width  is 

7T/72  Q 

4x-^-2  =  7Tp  X  40,000. 

Equating  these  two  results  and  solving  for  <?,  we  have 

(c  _  |)  _5_  x  60,000  =  TT  T9g  x  40,000, 
whence  c  =  4.5  in. 

As  a  check  on  the  correctness  of  our  assumptions  the  efficiency  of 
the  joint  is  found  to  be 

c-d     4.5  -.75 
e  =  -  =  -  :-=  —  =  .00. 
c  4.5 

92.  Structural  steel.  For  bridge  and  structural  work  the  following 
empirical  rules  are  representative  of  American  practice  :  * 

The  pitch  (or  distance  from  center  to  center)  of  rivets  should  not 
be  less  than  3  diameters  of  the  rivet.  In  bridge  work  the  pitch 
should  not  exceed  6  in.  or  16  times  the  thickness  of  the  thinnest 

*  Given  by  Cambria  Steel  Co. 


150  RESISTANCE  OF  MATERIALS 

outside  plates  except  in  special  cases  hereafter  noted.  In  the 
flanges  of  beams  and  girders,  where  plates  more  than  12  in.  wide 
are  used,  an  extra  line  of  rivets  with  a  pitch  not  greater  than  9  in. 
should  be  driven  along  each  edge  to  draw  the  plates  together. 

At  the  ends  of  compression  members  the  pitch  should  not  ex- 
ceed 4  diameters  of  the  rivet  for  a  length  equal  to  twice  the  width 
or  diameter  of  the  member. 

In  the  flanges  of  girders  and  chords  carrying  floors  the  pitch 
should  not  exceed  4  in. 

For  plates  in  compression  the  pitch  in  the  direction  of  the  line  of 
stress  should  not  exceed  16  times  the  thickness  of  the  plate,  and 
the  pitch  in  a  direction  at  right  angles  to  the  line  of  stress  should 
not  exceed  32  times  the  thickness,  except  for  cover  plates  of  top 
chords  and  end  posts,  in  which  the  pitch  should  not  exceed  40  times 
their  thickness. 

The  distance  between  the  edge  of  any  piece  and  the  center  of  the 
rivet  hole' should  not  be  less  than  1|  in.  for  |-in.  and  J-in.  rivets, 
except  in  bars  less  than  2|  in.  wide ;  when  practicable  it  should 
be  at  least  2  diameters  of  the  rivet  for  all  sizes,  and  should  not 
exceed  8  times  the  thickness  of  the  plate. 

Typical  illustrations  of  riveted  connections  in  structural  steel 
work  are  shown  in  Figs.  106  and  107. 

93.  Unit  stresses.  In  structural-steel  work  it  is  customary  to 
proportion  the  various  members  on  the  basis  of  certain  specified 
unit  stresses.  The  following  specifications  for  the  greatest  allow- 
able unit  stresses  represent  the  best  American  practice : 


Axial  tension,  net  section 

Axial  compression,  gross  section,  but  not  to  exceed  17,000 

lb./in.2 

I  =  length  of  member  in  inches, 
r  —  least  radius  of  gyration  of  member  in  inches. 
Compression  in  flanges  of  deck  plate  girders  and  built  or 

rolled  beams,  but  not  to  exceed  17,000  lb./in.2 

Compression  in  flanges  of  through  plate  girders,  but  not  to 
exceed  17,000  lb./in.2 


I  =  unsupported  length  of  flange  in  inches, 
r  =  least  radius  of  gyration  of  flange  section  laterally 
in  inches, 


18,000  lb./in.2 

16,000-17- 
r 


18,000  -  70  - 

18,000-  70- 

177)  W 


KIVETED  JOINTS  AND  CONNECTIONS 


151 


D  =  depth  from  top  of  girder  to  bottom  of  floor  beams, 
d  =  depth  of  floor  beams  back  to  back   of   angles   or 

flanges  of  beams, 
W '=  width  center  to  center  of  girders. 

Shear  in  webs  of  plate  girders,  net  section 

Shear  in  pins  and  shop-driven  rivets 

Shear  in  field-driven  rivets 

Tension  in  extreme  fiber  of  flanges  of  beams  proportioned 
by  moment  of  inertia,  net  section 

Tension  or  compression  in  the  extreme  fiber  of  pins,  assum- 
ing the  stresses  to  be  applied  in  the  centers  of  bearings    .    .    . 

Bearing  on  pins  in  members  not  subject  to  reversal  of  stress, 

Bearing  on  pins  in  members  subject  to  reversal  of  stress, 
using  the  greater  of  the  two  stresses 

Bearing  on  shop-driven  rivets  and  stiffeners  of  girders,  and 
other  parts  in  contact 

Bearing  on  concrete  masonry 

Bearing  on  sandstone  and  limestone  masonry 

Bearing  on  expansion  rollers  in  pounds   per   lineal   iiirh, 
where  d  =  diameter  of  roller  in  inches  . 


13,500  Ib./in.2 

13,500  Ib. /in. 2 
10,800  Ib./in.2 

18,000  Ib./in.2 

27,000  Ib./in.2 
24,000  Ib./in.2 

12,000  Ib./in.2 

27,000  Ib./in.2 
500  Ib. /in.2 
400  Ib./in.2 

GOO  d. 


APPLICATIONS 

257.  In  a  single-riveted  lap  joint  calculate  the  pitch  of  the  rivets  and  the  dis- 
tance from  the  center  of  the  rivets  to  the  edge  of  the  plate  under  the  assumption 
that  the  diameter  of  the  rivets  is  twice  as  great  as  the  thickness  of  the  plate. 

Solution.  Consider  a  strip  of  width  equal  to  the  rivet  pitch,  that  is,  a  strip  con- 
taining one  rivet.  Let  q  denote  the  unit  shearing  strength  of  the  rivet  and  p  the 
unit  tensile  strength  of  the  plate.  Then  if  h  denotes  the  thickness  of  the  plate,  in 
order  that  the  shearing  strength  of  the  rivet  may  be  equal  to  the  tensile  strength 
of  the  plate  along  the  line  of  rivet  holes,  we  must  have 

•rrd2 

—  q  =  (c-d)hp. 

Since  the  rivet  is  usually  of  better  material  than  the  plate,  we  may  assume  that  the 
ultimate  shearing  strength  of  the  rivet  is  equal  to  the  ultimate  tensile  strength 
of  the  plate  ;  that  is,  assume  that  p  —  q.  Under  this  assumption  the  above  relation 
becomes 


whence 


c  =  2.5d,  approximately. 


Similarly,  in  order  that  the  joint  may  be  equally  secure  against  shearing  off  the 
rivet  and  pulling  it  out  of  the  plate,  that  is,  shearing  the  plate  in  front  of  the  rivet, 
the  condition  is 


FIG.  106.    Detail  of  column  riveting  with  Bethlehem  I-beams  and  H  columns 


152 


^  i  i 

,  O       O  F1 

oojiq  1 1  q  '|oo 
oo]!"o~!|b  ijoo 

---O  |i  O  \c--'^ 

o  Ijoi     7 

OHO 


FIG.  107.    Riveted  joints  in  shop-building  construction  with  Bethlehem  wide-flange 
beams  used  for  columns  and  crane  girders 


153 


154  RESISTANCE  OF  MATERIALS 

where  a  denotes  the  margin  or  distance  from  center  of  rivets  to  edge  of  plate, 
and  q'  denotes  the  ultimate  shearing  strength  of  the  plate.   Assuming  that  q'  =  |  q 

and  h  =  -  ,  and  solving  the  resulting  expression  for  a,  we  have 

a  =  1.5d. 

258.  Find  the  required  rivet  pitch  for  a  single-riveted  lap  joint  with  l-in.  steel 
plates  and  J-in.  steel  rivets,  in  order  that  the  joint  shall  be  of  equal  strength  in 
shear  and  tension.  2 

Solution.  For  a  strip  of  length  c  the  strength  in  shear  is  —  -  g,  and  in  tension 
is  (c  —  d)hp.  Hence,  to  satisfy  the  given  condition, 

—q, 


Also,  efficiency  of  joint  is  then 

sectional  area  through  rivet  holes      c  —  d       .75 
e=  -  —  =  50  per  cent. 

sectional  area  unriveted  plate  c          1.50 

259.  Determine  the  maximum  diameter  of  rivet  in  terms  of  thickness  of  plate  so 
that  the  crushing  strength  of  the  joint  shall  not  be  less  than  its  shearing  strength. 

Solution.  It  is  customary  to  assume  the  ultimate  crushing  strength  of  the  material 
as  about  twice  its  ultimate  shearing  strength,  or  say  100,000  lb./in.2  Let 

pc  =  ultimate  crushing  strength, 
pt  =  ultimate  tensile  strength, 
q  =  ultimate  shearing  strength. 

Then,  if  the  shearing  strength  of  the  rivets  is  to  equal  their  crushing  strength,  we 
have  for  lap  joints 

—  q  -  dhpc          and         pc  =  2  q  ; 
whence  d  =  2.54  h. 

A  larger  rivet  than  this  will  crush  before  it  shears,  whereas  a  smaller  one  will 
shear  before  it  crushes.  Therefore  the  crushing  strength  of  a  lap  joint  need  not 
be  considered  when  d  <  2.54  A,  as  it  usually  is. 

For  a  rivet  in  double  shear,  assuming  that  the  strength  of  the  rivet  in  double 
shear  is  twice  that  of  the  same  rivet  in  single  shear,*  we  have  for  butt  joints 

•jrd2 

2  .  —  —  q  =  dhpc         and        pc  =  2  q  ; 
4 

whence  d  =  1.273ft. 

If  the  diameter  exceeds  this  value,  the  rivet  will  fail  by  crushing,  whereas  if  it  is 
smaller,  it  will  fail  by  shear.  Consequently  the  crushing  strength  of  a  butt  joint 
need  not  be  considered  when  d  <  1.273  h. 

*  This  seems  to  be  substantiated  by  experiment,  although  the  English  Board  of  Trade 
specifies  that  a  rivet  in  double  shear  shall  be  assumed  to  be  only  1.75  times  as  strong  as 
if  in  single  shear. 


EIVETED  JOINTS  AND  CONNECTIONS  155 

260.  Design  a  double-riveted  butt  joint  to  have  an  efficiency  of  75  per  cent, 
using  1-in.  steel  plates  and  steel  rivets. 

Solution.  For  any  strip  along  the  joint  of  length  equal  to  the  pitch  c,  two  rivets 
are  in  double  shear.  Hence  for  equal  strength  in  tension  and  shear  we  have 


Also,  —    —  =  e,  whence  d  =  c  (I  —  c)  and  c  —  d  =  ce.    Hence,  substituting   these 
values  of  c  —  d  and  d  in  the  first  equation,  the  result  is 

cchp  -  7rc2(l—  c)'2q; 

whence  c  =  —    —  -  -  --  =  4.586  in., 

7T(l-e)22 

or  say  c  =  4%  in. 

Then  d  =  c  (1  —  e)  =  1.150,     or  say  d  =  l^V  in. 

From  the  results  of  problem  259  it  is  apparent  that  the  crushing  strength  of  the 
joint  need  not  be  considered,  since  here  d  <  1.273/i. 

The  butt  straps  should  apparently  each  be  half  as  thick  as  the  plate,  but  when 
so  designed  they  are  found  to  be  the  weakest  part  of  the  joint.  It  is  therefore 
customary  to  make  the  thickness  of  the  butt  straps  about  |  A,  or,  in  the  present 
case,  |  in. 

261.  In  a  double-riveted  lap  joint  the  plates  are   ^  in.  thick,  rivets  ?  in.  in 
diameter,  and  pitch  3  in.    Calculate  the  efficiencies  of  the  joint  and  determine 
how  it  will  fail. 

262.  A  boiler  shell  is  to  be  4  ft.  in  diameter,  with  double-riveted  lap  joints,  and 
is  to  carry  a  steam  pressure  of  90  lb./in.2  with  a  factor  of  safety  of  5.    Determine 
the  thickness  of  shell  plates  and  diameter  and  pitch  of  rivets.    Also  calculate  the 
efficiency  of  the  joint. 

263.  A  cylindrical  standpipe  is  75  ft.  high  and  25  ft.   inside  diameter,  with 
double-riveted,  two-strap  butt  joints.    Determine  the  required  thickness  of  plates 
near  the  bottom  for  a  factor  of  safety  of  5,  and  also  the  diameter  and  pitch  of  rivets. 

264.  A  boiler  shell   ^  in.  thick  and  5ft.  in  diameter  has  longitudinal,  single- 
riveted  lap  joints,  with  1-in.  rivets  and  2^-in.  rivet  pitch.    Calculate  the  maximum 
steam  pressure  which  can  be  used  with  a  factor  of  safety  of  5. 

265.  A  cylindrical  standpipe  80  ft.  high  and  20  ft.  inside  diameter  is  made  of 
J-in.  plates  at  the  base,  with  longitudinal,  single-riveted,  two-strap  butt  joints, 
connected  by  1-in.  rivets  with  a  pitch  of  3J,  in.   Compute  the  factor  of  safety  when 
the  pipe  is  full  of  water. 

266.  Determine  the  diameter  and  pitch  of  rivets  required  to  give  the  strongest 
single-riveted  lap  joint,  using  |-in.  steel  plates  and  steel  rivets,  and  calculate  the 
efficiency  of  the  joint. 


SECTION  XIII 

REENFORCED  CONCRETE 

94.  Physical  properties.  The  use  of  concrete  dates  from  the 
time  of  the  Romans,  who  obtained  a  good  artificial  stone  from  a 
mixture  of  slaked  lime,  volcanic  dust,  sand,  and  broken  stone.  The 
modern  use  of  concrete,  however,  is  of  comparatively  recent  develop- 
ment, its  universal  use  being  a  matter  of  only  the  last  quarter  of  a 
century,  while  reenforced  concrete  is  of  still  more  recent  origin. 

Concrete  is  made  by  mixing  broken  stone,  varying  in  size  from 
a  walnut  to  a  hen's  egg,  with  clean,  coarse  sand  and  Portland  cement, 
using  enough  water  to  make  a  mixture  of  the  consistency  of  heavy 
cream.  The  proportion  of  these  three  materials  depends  on  their 
relative  size  ;  in  general,  enough  sand  being  needed  to  fill  the  voids 
in  the  broken  stone  and  enough  cement  to  fill  the  voids  in  the  sand. 
The  cement  and  water  cause  the  mass  to  begin  to  stiffen  in  about 
half  an  hour,  and  in  from  ten  to  twenty-four  hours  it  becomes  hard 
enough  to  resist  pressure  with  the  thumb.  In  a  month  the  mixture 
becomes  thoroughly  hard,  although  the  hardness  continues  gradually 
to  increase  for  some  time. 

Portland  cement  was  invented  by  Joseph  Aspdin  of  Leeds,  Eng- 
land, who  took  out  a  patent  for  its  manufacture  in  1824,  the  name 
Portland  being  due  to  its  resemblance  to  a  popular  limestone 
quarried  in  the  Isle  of  Portland.  Its  manufacture  was  begun  in 
1825,  but  its  use  did  not  become  general  until  1850,  when  the 
French  and  the  Germans  became  active  in  its  scientific  production 
and  succeeded  in  greatly  improving  both  the  method  of  manu- 
facture and  the  quality  of  the  finished  product.  Portland  cement 
was  first  brought  to  the  United  States  in  1865,  but  not  until  1896 
did  its  annual  domestic  production  reach  a  million  barrels. 

When  ordinary  limestone  (calcium  carbonate)  is  heated  to  about 
800°  F.,  carbon  dioxide  is  driven  off,  leaving  an  oxide  of  calcium 
called  quicklime.  This  has  a  great  affinity  for  water,  and  when 

156 


REENFORCED  CONCRETE  157 

combined  with  it  is  said  to  be  slaked.  Slaked  lime  when  dry  falls 
into  a  fine  powder. 

Lime  mortar  is  formed  by  mixing  slaked  lime  with  a  large  pro- 
portion of  sand.  Upon  exposure  to  the  air  this  mortar  becomes 
hard  by  reason  of  the  lime  combining  with  carbon  dioxide  and 
forming  again  calcium  carbonate,  the  product  being  a  sandy  lime- 
stone. Lime  mortar  is  used  in  laying  brick  walls  and  in  structures 
where  the  mortar  will  not  be  exposed  to  water,  since  it  will  not  set, 
that  is,  combine  with  carbon  dioxide,  under  water. 

When  limestone  contains  a  considerable  amount  of  clay,  the  lime 
produced  is  called  hydraulic  lime,  for  the  reason  that  mortar  made 
by  using  it  will  harden  under  water.  If  the  limestone  contains 
about  30  per  cent  of  clay  and  is  heated  to  1000°  F.,  the  carbon 
dioxide  is  driven  off,  and  the  resulting  product,  when  finely  ground, 
is  called  natural  cement.  When  about  25  per  cent  of  water  is  added, 
this  cement  hardens  because  of  the  formation  of  crystals  of  calcium 
and  aluminium  compounds. 

If  limestone  and  clay  are  mixed  in  the  proper  proportions,  usually 
about  three  parts  of  lime  carbonate  to  one  of  clay,  and  the  mixture 
roasted  to  a  clinker  by  raising  it  to  a  temperature  approaching 
3000  F.,  the  product,  when  ground  to  a  fine  powder,  is  known  as 
Portland  cement.  The  proper  proportion  of  limestone  and  clay  is 
determined  by  finding  the  proportions  of  the  particular  clay  and 
stone  that  will  make  perfect  crystallization  possible.  In  the  case  of 
natural  cement  the  lime  and  clay  are  not  present  in  such  propor- 
tions as  to  form  perfect  crystals,  and  consequently  it  is  not  as  strong 
as  Portland  cement. 

The  artificial  mixing  of  the  limestone  and  clay  in  the  manufac- 
ture of  Portland  cement  is  accomplished  in  different  ways.  Through- 
out the  north  central  portion  of  the  United  States  large  beds  of  marl 
are  found,  and  also  in  the  same  localities  beds  of  suitable  clay. 
This  marl  is  nearly  pure  limestone  and  is  mixed  with  the  clay 
when  wet.  (These  materials  are  also  mixed  dry.)  Both  the  marl 
and  clay  are  pumped  to  the  mixer,  where  they  are  mixed  in  the 
proper  proportions.  The  product  is  then  dried,  roasted,  and  ground. 
Most  American  Portland  cements,  however,  are  made  by  grinding 
a  clay-bearing  limestone  with  sufficient  pure  limestone  to  give  the 


158  RESISTANCE  OF  MATERIALS 

proper  proportions.  After  being  thoroughly  mixed,  the  product  is 
roasted  and  ground  to  a  powder. 

Slag  cement  (puzzolan)  is  made  by  thoroughly  mixing  with  slaked 
lime  the  granulated  slag  from  an  iron  blast  furnace  and  then  grind- 
ing the  mixture  to  a  fine  powder.  Slag  cements  are  usually  lighter 
in  color  than  the  Portland  cements  and  have  a  lower  specific  gravity, 
the  latter  ranging  from  2.7  to  2.8.  They  are  also  somewhat  slower 
in  setting  than  the  Portland  cements  and  have  a  slightly  lower 
tensile  strength.  They  are  not  adapted  to  resist  mechanical  wear, 
such  as  would  be  necessary  in  pavements  and  floors,  but  are  suitable 
for  foundations  or  any  work  not  exposed  to  dry  air  or  great  strain. 

True  Portland  cement  may  be  made  from  a  mixture  of  blast- 
furnace slag  and  finely  powdered  limestone,  the  mixture  being 
burned  in  a  kiln  and  the  resultant  clinker  ground  to  powder.  Both 
the  Portland  and  the  puzzolan  cements  will  set  under  water; 
that  is,  they  are  hydraulic. 

Gravel  or  broken  stone  forms  the  largest  part  of  the  mass  of  a 
good  concrete  and  is  called  the  coarse  aggregate.  Its  particles 
may  be  from  •£-  in.  to  |-  in.  in  diameter  for  thin  walls  or  where 
reenforcement  is  used,  or  up  to  2i  in.  for  heavy  foundations  or 
walls  over  a  foot  thick.  The  coarse  aggregate  should  always  be 
clean  and  hard. 

The  sand,  or  fine  aggregate,  should  be  clean  and  coarse ;  that  is,  a 
large  proportion  of  the  grains  should  measure  ^  to  1  in.  in  diam- 
eter. All  should  pass  through  a  screen  of  i-in.  mesh.  Too  fine 
a  sand  weakens  the  mixture  and  requires  a  larger  proportion  of 
cement. 

The  following  standard  proportions  may  be  taken  as  a  guide  to 
the  proper  mixture  for  various  classes  of  work:* 

1.  A  rich  mixture  for  columns  and  other  structural  parts  sub- 
jected to  high  stresses  or  required  to  be  exceptionally  water-tight. 
Proportions  1 :  1|  :  3 ;  that  is,  one  barrel  (4  bags)  of  packed  Portland 
cement  to  one  and  one  half  barrels  (5.7  cu.  ft.)  of  loose  sand  to 
three  barrels  (11.4  cu.  ft.)  of  loose  gravel  or  broken  stone. 

*  Taylor  and  Thompson,  "  Concrete  Plain  and  Reinforced  "  ;  also,  Atlas  Portland  Ce- 
ment Co.,  "  Concrete  Construction,"  and  Turneaure  and  Manrer,  "  Principles  of  Rein- 
forced Concrete  Construction,"  p.  10. 


REENFORCED  CONCRETE  159 

2.  A  standard  mixture  for  reenforced  floors,  beams,  and  columns, 
for  arches,  for  reenforced  engine  or  machine  foundations  subject  to 
vibrations,  and  for  tanks,  sewers,  conduits,  and  other  water-tight 
work.    Proportions  1:2:4;  that  is,  one  barrel  (4  bags)  of  packed 
Portland  cement  to  two  barrels  (7.6  cu.  ft.)  of  loose  sand  to  four 
barrels  (15.2  cu.  ft.)  of  loose  gravel  or  broken  stone. 

3.  A  medium    mixture   for    ordinary   machine    foundations,   re- 
taining  walls,   abutments,   piers,   thin   foundation  walls,   building 
walls,  ordinary  floors,  sidewalks,  and  sewers  with  heavy  walls.    Pro- 
portions 1  :  2J  :  5  ;  that  is,  one  barrel  (4  bags)  of  packed  Portland 
cement  to  two  and  one  half  barrels  (9»5  cu.  ft.)  of  loose  sand  to 
five  barrels  (19  cu.  ft.)  of  loose  gravel  or  broken  stone. 

4.  A  lean  mixture  for  unimportant  work  in  masses,  for  heavy 
walls,  for  large  foundations  supporting  a  stationary  load,  and  for 
backing  for  stone  masonry.   Proportions  1:3:6;  that  is,  one  barrel 
(4  bags)  of  packed  Portland  cement  to  three  barrels  (11.4  cu.  ft.)  of 
loose  sand  to  six  barrels  (22.8  cu.  ft.)  of  loose  gravel  or  broken  stone. 

95.  Design  of  reenforced  concrete  beams.  Since  concrete  is  a  mate- 
rial which  does  not  conform  to  Hooke's  law  and  moreover  does 
not  obey  the  same  elastic  law  for  tension  as  for  compression,  the 
exact  analysis  of  stress  in  a  plain  or  reenforced  concrete  beam  would 
be  much  more  complicated  than  that  obtained  under  the  assump- 
tions of  the  common  theory  of  flexure.  The  physical  properties  of 
concrete,  however,  depend  so  largely  on  the  quality  of  material  and 
workmanship,  that  for  practical  purposes  the  conditions  do  not  war- 
rant a  rigorous  analysis.  The  following  simple  formulas,  although 
based  on  approximate  assumptions,  give  results  which  agree  closely 
with  experiment  and  practice. 

Consider  first  a  plain  concrete  beam,  that  is,  one  without  reen- 
forcement.  The  elastic  law  for  tension  is  in  this  case  (see  Fig.  108) 


and  for  compression  ^-£—  =  Ec. 

Sc 

To  simplify  the  solution,  however,  assume  the  straight-line  law  of 
distribution  of  stress  ;  that  is,  assume  ml  =  m2  =  1.    Note,  however, 


160 


RESISTANCE  OF  MATERIALS 


that  this  does  not  make  the  moduli  equal.  Assume  also  that  cross 
sections  which  were  plane  before  flexure  remain  plane  after  flexure 
(Bernoulli's  assumption),  which  leads  to  the  relation 


where  ec  and  et  denote  the  distances  of  the  extreme  fibers  from  the 
neutral  axis  (Fig.  108). 

Now  let  the  ratio  of  the  two  moduli  be  denoted  by  n ;  that  is,  let 


it 


Then  £s  =  ^s  =  w^. 

Pt      *&         et 

For  a  section  of  unit  width  the  resultant 
compressive  stress  Rc  on  the  section  is 
Rc  =  \pcec,  and  similarly  the  resultant 
tensile  stress  Rt  is  Bt=±ptet.  Also,  since 

^  Rc  and  Rt  form  a  couple,  Rc  =  Rt.  Hence 

FIG.  108  v       e 

pcec  =  ptet,  or  •&•  =  — »  and,   equating  this 

v  ^t      €° 

to  the  value  of    the    ratio  —    obtained   above,  we    have 

Pt 


Since  the  total  depth  of  the  beam  h  is  h  =  ec  4-  ep  we  have,  therefore, 
ec  =  h  —  ec  Vw,  whence  , 


and   similarly   et  =  h 'j=  ,  whence 

h 

^^ 

Vw 

Now,  by  equating  the  external  moment  Mto  the  moment  of  the 
stress  couple,  we  have 


or   ^= 


REENFORCED   CONCRETE 
whence,  by  solving  for  the  unit  stresses  pc  and  pt 


161 


7 
n 


or,  solving  one  of  these  two  relations  for  A,  say  the  first,  we  have 


7  ItJ  -LVI.  /_,  /       \ 

^  =  ^— (1  +  Vra). 

For  ordinary  concrete  ^  may  be  taken  as  25.  Also,  using  a  factor 
of  safety  of  8,  the  working  stress pc  becomes pc  =  300  lb./in.2  Substi- 
tuting these  numerical  values  in  the  above,  the  formula  for  the  depth 
of  the  beam  in  terms  of  the  external  moment  takes  the  simple  form 


FIG.  109 


h  = 

4 

h  being  expressed  in  inches,  and  M  in 
inch-pounds  per  inch  of  width  of  beam. 
For  a  reenforced  concrete  beam  the 
tensile  strength  of  the  concrete  may  be 
neglected.  Let  Ec  and  Es  denote  the  moduli  of  elasticity  for  con- 

771 

crete  and  steel  respectively,  and  let  —£  =  n.    Then,  if  x  denotes  the 

distance  of   the  neutral  axis   from  the  top  fiber  (Fig.   109),  the 
assumptions  in  this  case  are  expressed  by  the  relations 


h  —  x 


and     —  = 


whence 


or,  solving  for  x,  x—li — — 

P.  +  nPc 

Now  if  A  denotes  the  area  of  steel  reenforcement  per  unit  width  of 

beam,  then 

Es  =  p8A     and  Rc  =  \pcx ; 

and  consequently,  since  Rc  =  JBg, 


162  KESISTANCE  OF  MATEEIALS 

Moreover,  equating  the  external  moment  M  to  the  moment  of  the 
stress  couple,  we  have 


Substituting  the  value  of  x  in  either  one  of  these  expressions,  say 
the  first,  we  have 


whence,  solving  for 


For  practical  work  assume  w  =  15,  pc=  500  lb./in.2  (factor  of 
safety  of  5),  and  ps  =  15,000  lb./in.2  (factor  of  safety  of  4). 
Substituting  these  numerical  values  in  the  above,  the  results  take 
the  simple  form 


h 

A  =  -  ,  x  =  6O  A, 

180 


where  H  denotes  the  total  depth  of  the  beam  in  inches,  d  is  the 
diameter  of  the  reenforcement  in  inches,  and  M  is  the  external 
moment  in  inch-pounds  per  inch  of  width. 

In  designing  beams  by  these  formulas  first  find  A,  then  A,  and 
finally  H. 

96.  Calculation  of  stirrups,  or  web  reenforcement.  For  a  beam 
reenforced  with  horizontal  rods  only,  that  is,  having  no  vertical  or 
web  reenforcement,  the  ultimate  shearing  strength  is  found  to  be 
about  100  lb./in.2,  calculated  as  the  average  shearing  stress  on  the 
cross  section.  The  working  stress  in  shear  for  the  concrete  is  there- 
fore assumed  to  be  25  or  30  lb./in.2,  equivalent  to  a  factor  of  safety 
of  3  or  4. 

If  the  average  shear  on  any  cross  section  exceeds  30  lb./in.2, 
vertical,  or  web,  reenforcement  is  required,  usually  supplied  in  the 


REENFORCED  CONCRETE 


163 


form  of  stirrups,  or  loops  (Fig.  110).  It  can  be  shown  that  the 
maximum  shear  in  a  beam  is  inclined  at  an  angle  of  45°  to  the  axis 
of  the  beam.*  Therefore  to  be  effective,  vertical  stirrups  cannot  be 
spaced  farther  apart  than  the  depth  of  the  beam.  In  actual  prac- 
tice it  is  customary  to  make  the  distance  apart  about  one  half  this 

amount,  or  - ,   where  h  denotes  the  depth  of  the  beam. 

Since  the  maximum  shear  is  inclined  at  an  angle  of  45°  to  the 
vertical,  the  effective  area  of  the  stirrups  is  V2  times  their  cross- 
sectional  area ;  but  since  the  maximum  shear  is  also  approximately 
equal  to  V2  times  the  average  shear, t  it  is  usual  simply  to  design 
the  stirrups  to  carry  the  average  shear. 


Longitudinal  Section 
Through  Beam  and  Columns 


Column 


Transverse  Section 
Through  Beams  and  Slab 

FIG. 110 

For  instance,  suppose  that  the  maximum  shear  on  any  cross  sec- 
tion of  a  beam  has  an  average  value  of  100  lb./in.2,  which,  as  shown 
by  the  results  of  tests,  is  about  the  maximum  limit  for  good  work. 
Then,  assuming  that  the  concrete  carries  30  lb./in.2  of  this  shear, 
the  vertical  stirrups  must  be  designed  to  carry  the  remainder,  or 
70  lb./in.2  Therefore  if  the  beam  is  of  breadth  b  and  depth  A,  the 
total  shear  to  be  carried  by  the  stirrups  is  70  bh,  and  consequently  for 
a  working  stress  of  15,000  lb./in.2  in  the  steel,  the  required  area  is 

70  bh 


15,000 

or  .47  per  cent  of  the  total  area  of  the  cross  section.     Since  the 
stirrups  are  usually  in  the  form   of   a   double  loop,  the  required 

*  Slocum  and  Hancock,  Strength  of  Materials,  revised  edition,  article  28,  p.  25,  Ginn 
and  Company.  t  Ibid.,  article  55,  pp.  59  and  60. 


164  BESISTANCE  OF  MATERIALS 

cross-sectional  area  of  each  stirrup  rod  is  .23  per  cent  of  the  total 
area  of  the  cross  section. 

Inclined  reenf  orcing  rods,  formed  by  bending  up  part  of  the  hori- 
zontal bottom  rods  at  an  angle  of  45°,  are  usually  too  large  and  too 
far  apart  to  form  an  effective  web  reenf orcement,  but  a  combination 
of  the  two,  as  shown  in  Fig.  110,  constitutes  the  most  effective  design. 

97.  Reenf orced  concrete  columns.  It  is  seldom  necessary  to  design 
reenforced  concrete  columns  by  the  formulas  for  long  columns.  In 
ordinary  construction  the  ratio  of  length  to  least  width  seldom  ex- 
ceeds 12  or  15,  while  actual  tests  show  that  they  may  be  practically 
considered  as  short  blocks  for  ratios  up  to  20  or  25.  The  strength 
of  a  reenforced  concrete  column  considered  as  a  short  block  will 
therefore  first  be  determined,  and  in  exceptional  cases  this  result 
may  then  be  corrected  by  applying  a  general  column  formula. 

The  method  of  reenforcing  concrete  columns  is  either 

1.  by  means  of  longitudinal  rods  extending  the  full  length  of 
the  column ; 

2.  by  means  of  hoops  or  spiral  bands ; 

3.  by  a  combination  of  longitudinal  rods  and  hoops  or  spirals. 
Let  A  denote  the  total  cross-sectional  area  of  the  column  ;  Ae  the 

area  of  the  concrete  ;  As  the  area  of  the  steel ;  and  pc,  ps  the  safe  unit 
stresses  in  the  concrete  and  steel  respectively.  Then  the  safe  load 
P  for  the  column  is  given  by 

P=PrAc+psAs. 

The  unit  deformations  of  the  concrete  and  steel  corresponding  to 
these  stresses  are 

•-!•  -I- 

where  Ec  and  Es  denote  Young's  moduli  for  the  concrete  and  steel 
respectively.  Since  the  concrete  and  steel  must  deform  the  same 
amount,  sc  =  *,,  and,  consequently, 

Ps          ES 

=  lf  =  n> 

PC         EC 

or  ps  =  npc, 

where  n  denotes  the  ratio  of  the  i^ro  moduli,  ordinarily  assumed 
to  be  15. 


REENFORCED  CONCRETE  165 

It  is  desirable  to  express  the  load  P  in  terms  of  the  total  area  of 
the  cross  section  A.  For  this  purpose  let  Jc  denote  the  percentage 
of  reenforcemeiit,  or  the  ratio  of  the  area  of  the  steel  to  the  total 
area  ;  that  is,  let 


Then  Ac  =  A  -  As  =  A-  JcA  =  A(l-  F). 

Therefore          P  =  pcA(.  +  p^As  =  pcA  (1  —  k)  -f  npJcA  ; 
whence  1>  =  pcA  [1  +  (n  -  1)  k]  . 

If  the  column  was  plain  concrete  without  reinforcement,  its  safe 
load  would  be  P'  =  pcA.  The  relative  strength  of  a  plain  concrete 
column  as  compared  with  one  reenforced  is  therefore 


Thus,  if  k  =  1  per  cent  and  n  =  15,  we  have 


that  is,  a  reenforcemeiit  of  1  per  cent  of  metal  increases  the  strength 
14  per  cent. 

In  the  case  of  reenforcemeiit  in  the  form  of  hoops  or  spirals,  the 
increase  in  strength  depends  on  the  effect  of  the  hoops  or  coils  in 
preventing  lateral  deformation.  The  results  of  tests  show  that  this 
effect  is  very  slight  for  loads  up  to  the  ultimate  strength  of  plain 
concrete,  but  beyond  this  point  there  is  a  notable  increase  in  the 
ultimate  strength  of  the  column.  Tests  on  hooped  columns  made 
under  the  direction  of  Professor  A.  N.  Talbot  at  the  University  of 
Illinois  showed  that  the  ultimate  strength  of  the  column  in  terms 
of  the  percentage  of  steel  reeiiforcement  may  be  calculated  by 
the  formulas 

for  mild  steel,  p  =  1600  +  65,000  k, 

for  high  steel,  p  =  1600  -f  100,000  Jc  ; 

where  p  denotes  the  stress  per  square  inch,  and  k  is  the  percentage 
of  steel  with  reference  to  the  concrete  core  inside  the  hoops.  The 
compressive  strength  of  plai»  concrete  is  here  assumed  to  be 
1600  lb./in.2  As  regards  ultimate  strength,  the  effect  of  the 


166  RESISTANCE  OF  MATERIALS 

hoop  reenforcement  was  found  to  be  from  2  to  4  times  as  great 
as  for  the  same  amount  of  metal  in  the  form  of  longitudinal  rods. 
From  extensive  investigations  and  experiments  on  hooped  col- 
umns, Considere  has  derived  the  formula 


where  P  =  ultimate  strength  of  the  column, 

pc  =  ultimate  strength  of  the  concrete, 
ps  =  elastic  limit  of  the  steel. 

This  formula  indicates  that  hoops  or  spirals  are  2.4  times  as  effec- 
tive as  the  same  amount  of  metal  in  the  form  of  longitudinal  rods. 
When  longitudinal  rods  are  used  without  hoops,  it  is  necessary 
to  tie  them  together  at  intervals  to  prevent  them  from  buckling 
and  pulling  away  from  the  concrete.  The  distance  apart  for  these 
horizontal  ties  may  be  determined  by  considering  the  longitudinal 
reenf  orcing  rods  as  long  columns  and  applying  Euler's  formula  ; 
namely,  ^EI 

~lT 

Assuming  a  factor  of  safety  of  5,  and  taking  E=  25,000,000  lb./in.2 
for  wrought  iron,  we  have 


10  x  25,000,000  x 
2  _  T^EI  _  _  64 

5P  ~      Trd'2 


where  d  denotes  the  diameter  of  the  reenforcing  rods,  from  which 
the  unsupported  length  I  of  the  rods,  or  distance  between  ties,  is 
found  to  be  _1750f* 


For  instance,  suppose  that  a  concrete  column  12  in.  square,  carry- 
ing a  load  of  80,000  lb.,  is  reenforced  with  four  rods  |  in.  in  diam- 
eter placed  in  the  four  corners,  1  in.  from  the  outer  faces.    Then 
A  =  144  in.2,     As  =  4  (.6013)  =  2.4052  in.2, 


REENFORCED  CONCRETE 


167 


Therefore  the  distance  between  ties  should  be 

1750- 


-18  in. 


98.  Radially  reenforced  flat  slabs.  A  system  of  floor  construction 
without  the  use  of  beams  or  ribs,  called  the  "  mushroom  system," 
has  been  devised  by  Mr.  C.  A.  P.  Turner  (Fig.  111).  The  essen- 
tial features  of  this  system  are  that  the  floor  slab  is  of  uniform 
thickness  throughout,  the  reenforcement  is  radial,  and  the  col- 
umn top  is  enlarged  and  reenforced  with  hoops.  This  type  of  con- 
struction is  best  adapted  to  large  areas  with  few  large  openings. 
The  following  is  a  simple  analysis 
of  the  chief  features  of  the  design.  f*—  —-D- 


(PI  n) 


(Vertical  If  ||  Section) 


FIG. Ill 


99.  Diameter  of  top.  In  the  case  of  a  continuous  floor  slab  sup- 
ported by  several  columns,  it  is  obvious  that  the  slab  will  be  con- 
cave downwards  over  the  columns  and  concave  upward  in  the  center 
of  each  panel.  Between  these  two  extremes  there  must  be  a  bound- 
ary at  which  there  is  no  curvature,  that  is,  a  line  of  inflection.  In 
a  restrained  beam  of  length  /,  bearing  a  uniform  load,  the  two 
points  of  inflection  occur  at  a  distance  of  0.212  I  from  each  support 
(article  45).  For  a  continuous  slab,  therefore,  the  line  of  inflection 
for  square  panels  may  be  assumed  to  be  approximately  a  circle  of 
radius  between  -£  I  and  ^  Z,  where  I  denotes  the  span,  or  distance 


168 


RESISTANCE  OF  MATERIALS 


center  to  center  of  columns.    For  practical  purposes  the  diameter 
D  of  the  column  top  may  therefore  be  assumed  as 


which  is  approximately  the  mean  of  the  above  values. 

There  is  another  condition,  however,  which  also  affects  the 
diameter  of  the  top,  namely,  the  distribution  of  slab  reeiiforce- 
ment.  If  the  column  top  is  too  small,  there  will  be  portions  of  the 
slab  which  contain  no  reeiiforcement,  as  shown  by  the  triangular 

areas  a,  6,  c,  d  (Fig.  112). 
The  arrangement  shown  in 
Fig.  113,  however,  has  no 
such  gaps  ;  it  requires  that 

D2  +  D2  =  (I  -  D)2, 


o] 


o 


:o 


:o 


FIG. 112 


FIG. 113 


or    D  = 


=  =  0.414  Z, 


which  determines  the  minimum  diameter  of  column  top.    If,  then, 
D  is  assumed  as 

L>  =  —l  =  0.4375^, 
16 

a  slight  overlap  of  the  reenforcing  rods  is  assured. 

100.  Efficiency  of  the  spider  hoops.  The  efficiency  of  tensile  re- 
enforcement  in  the  form  of  hoops  as  compared  with  direct  reen- 
forcement  may  be  obtained  approximately  as  follows : 

In  the  case  of  column  tops  as  here  considered,  take  a  diametral 
section  of  the  top  and  consider  half  of  one  hoop  and  the  portion  of 
the  material  reenforced  by  this  segment,  as  shown  in  Fig.  114.  Let 
w  denote  the  radial  stress,  or  pressure  on  the  inside  of  the  hoop 
per  unit  of  length,  expressed  in  pounds  per  linear  inch  of  hoop. 
Also  let  r  denote  the  radius  of  the  hoop,  A  its  cross-sectional  area, 
and  p  the  unit  stress  in  the  metal.  Then  the  radial  force  acting  on 
any  portion  of  the  hoop  of  length  As  is  wAs,  and  the  component  of 
this  force  perpendicular  to  the  plane  of  the  section  is  wAs  sin  a. 
Or,  if  A#  denotes  the  projection  of  As  on  the  diameter,  then 
As  sin  a  =  A#,  and  this  component  of  the  force  therefore  becomes 


BEENFORCED  CONCRETE 


169 


'w&x.  Therefore,  equating  the  tension  in  the  hoop  to  the  sum  of 
the  components  of  the  radial  stress  perpendicular  to  the  plane  of 
the  section,  we  have 


whence 


wr 


P 


Now   let  A'  denote   the   cross-sectional   area   of    an   equivalent 
amount  of  radial  reenforcement.    Then,   since  the  length  of  the 
arc  considered  is  TTT  and  the  radial  stress  is  of  amount  w  per  unit 
of  length,  the  total  amount 
of  radial  reenforcement  re- 
quired would  be  given  by 
the  equation 

pA'  =  TTTW  ; 


whence      A1  = 


7TTW 

P 


Comparing  these  expres- 
sions for  A  and  J/,  it  is  found 
that  A  i  _ 


FIG. 114 


Consequently,  the  theoretical  efficiency  of  tensile  reenforcement 
in  the  form  of  hoops  is  3.14  times  as  great  as  the  same  cross- 
sectional  area  of  direct,  or  radial,  reenforcement.  The  amount  of 
metal  in  a  hoop  of  radius  r,  however,  is  2  irrA,  whereas  that  in  the 
radial  reenforcement  is  Ar  2  r,  and  since  Ar  —  irA,  these  volumes  are 
equal.  Consequently,  there  is  no  saving  in  material  effected  by 
making  the  reenforcement  in  the  form  of  hoops.  But  when  there 
is  such  a  complex  system  of  reenforcement  as  that  shown  in 
Fig.  Ill,  some  of  the  metal  may  be  used  to  better  advantage  in 
the  form  of  hoops,  as  this  lessens  somewhat  the  congestion  of  metal 
at  the  columns. 

101.  Maximum  moment.  For  a  continuous  beam  of  span  Z, 
carrying  a  total  uniform  load  of  amount  W,  the  moment  at  the 

Wl 
supports  is  — — ;  whereas  the  moment  at  midspan  is  one  half  this 


170  RESISTANCE  OF  MATERIALS 

Wl 

amount,  or  —  —  .    Assuming  that  each  of  the  four  sets  of  slab  rods 
2i<± 

carries  one  fourth  the  total  load,  the  bending  moment  from  which 
to  determine  the  thickness  of  the  slab  and  the  amount  of  reenf  orce- 
ment  in  the  head  becomes 


Wl 


where  W  denotes  the  total  load  on  the  panel,  and  I  is  the  distance 
center  to  center  of  columns.    The  formula  determined  by  experi- 

Wl  7 

ment  and  used  in  practice  is  M  =  —  .  For  a  top  of  diameter  D  =  —  Z, 

oO  16 

Wl      7 

the  moment  per  foot  of  width,  say  Jtf  ,  is  therefore  M  =  —  -s-  —  Z, 

50      16 

Wl       7  7      W  .    ,, 
Jf='=ft-lb- 


102.  Thickness  of  slab.    Let 

ps  —  unit  working  tensile  stress  in  reenforcement, 
pc  =  unit  working  compressive  stress  in  concrete, 

7? 

n  =  —j-  =  ratio  of  elastic  moduli, 
Ml  =  bending  moment  in  foot-pounds  per  foot  of  width. 

Then  the  thickness  of  slab  h  from  the  outer  fiber  in  compres- 
sion to  the  center  of  the  reenforcement  is  given  by  the  formula 
(article  95) 


For  working  values  of  pg  =  16,000  lb./in.2, pc  =  600  lb./in.2,  and 

n  =  15  this  becomes 

h=  0.1026  Vjfr 

W 

Since  Ml  =  — ,  as  explained  above,  this  may  also  be  written 
22 

h=  0.02187  VJF, 

where  h  is  expressed  in  inches  and  W  denotes  the  total  load  on  the 
panel  in  pounds. 


KEENFORCED  CONCRETE  171 

The  total  thickness  of  slab  is  then  found  by  adding  to  this  value 
of  h  the  amount  needed  for  the  reenf  orcing  rods  plus  a  small  amount 
for  bond  below  the  bottom  rods. 

The  amount  of  reenforcement  for  the  unit  stresses  assumed  above 
is  then  found  from  the  relation  (article  95) 

A  =  0.081  A, 

where  h  is  expressed  in  inches  and  A  in  square  inches  per  foot 
of  width. 

103.  Area  of  slab  rods.    As  stated  above,  the  moment  at  the 
center  of  the  slab  is  half  as  great  as  at  the  supports.    The  effect  of 
this  on  the  required  dimensions  of  the  slab  and  reenf  orcement  would 
be  to  divide  both  h  and  A  by  V2.    But  since  the  slab  is  necessarily 
of  the  same  thickness  throughout  and  hence  is  thicker  at  the  center 
than  necessary,  the  cross-sectional  area  of  the  reenforcement  in  the 
slab  may  be  lessened,  so  as  to  make  the  moment  of  the  stress  in  the 
reenforcement  equal  to  the  moment  required  for  the  thinner  slab. 
If,  then,  p  denotes  the  unit  stress  in  the  metal,  where  p  is  assumed 
to  be  the  same  in  both  cases,  and  A'  denotes  the  cross-sectional  area 
of  metal  actually  required,  the  condition  that  the  moment  shall  be 
constant  is  , 

kpA'li  =  kp  ——  •  — — ; 

rV2    V2 

whence  A'  =  — . 

Consequently,  the  design  is  made  up  by  placing  half  the  required 
cross-sectional  area,  obtained  from  the  formula  A  =  0.081  A,  in 
the  slab  rods  and  the  other  half  in  the  hoops  and  spider  in  the 
column  top. 

104.  Application   of   formulas.    To  illustrate  the  use   of  these 
formulas,  consider  a  floor  system  with  panels  20  ft.  square,  carry- 
ing a  live  load  of  200  lb./ft.2    By  a  preliminary  calculation  it  is 
found  that  the  floor  slab  will  be  about  9  in.  thick,  giving  a  dead  load 
of  115  lb./ft.2   The  total  live  and  dead  load  is  therefore  315  lb./ft.2 
Consequently,  W=W  x  315-126,000  lb.,  and 


h  =  0.02187  Vl26,000  -  7.76  in. 


172  RESISTANCE  OF  MATERIALS 

The  total  area  of  reinforcement  in  the  head  for  each  radial  sys- 
tem is  then  A  =  0>ogl  h  =  Q>628  sq>  in>  per  foot 

Since  the  diameter  of  the  top  is  assumed  as  D  =  -^  1=  8.75  ft., 
the  total  area  required  for  one  radial  system  is  8.75^4  =  5.5  sq.  in. 
The  required  area  of  slab  rods  in  one  system  is  then 

-^-=  2.75sq.  in., 

equivalent  to  18  rods  T7g  in.  in  diameter,  spaced  5  in.  apart. 

Since  the  four  sets  of  rods  overlap  where  they  cross  the  column 
top,  and  since  h  denotes  the  distance  from  the  extreme  fiber  in  com- 
pression to  the  center  of  the  reenforcement,  the  total  thickness  of^ 
slab  becomes  h  +  2  d  4-  |-  in.  =  9  in. 

Since  the  hoops  around  the  column  head  are  assumed  to  be  TT 
times  as  effective  as  the  same  cross-sectional  area  of  radial  reen- 
forcement, the  total  area  of  the  hoops,  neglecting  the  spider,  is 

2  75 

— YJ=  0.88  sq.  in.    If  two  hoops  are  used,  the  diameter  of  each 

hoop  rod  may  therefore  be  assumed  as  |  in.,  giving  a  total  cross- 
sectional  area  of  0.88  sq.  in. 

It  should  be  noted,  however,  that  the  effectiveness  of  hoop  reen- 
forcement depends  on  the  hoop  being  placed  where  it  can  carry  the 
tensile  stress  in  the  column  top.  Since  the  outermost  hoop  of  the 
top  is  placed  as  nearly  as  possible  on  the  line  of  inflection,  there  is 
practically  no  stress  in  the  slab  at  this  point,  except  shear,  and 
hence  the  cross-sectional  area  of  the  outer  hoop  should  be  neglected 
in  dimensioning  the  top  hoops. 

105.  Dimension  table.  By  the  use  of  the  above  formulas  as  just 
explained,  the  accompanying  table  has  been  calculated,  giving  the 
required  thickness  of  slab  and  also  the  size  and  number  of  slab 
rods  for  various  spans  and  loads. 

In  this  table  the  thickness  of  concrete  below  the  bottom  of  the 
rods  has  been  assumed  as  about  i  in.,  which  has  been  proved  by 
experiment  to  be  sufficient  for  fireproofing  purposes.  Some  fire- 
proofing  specifications  require  more,  however,  in  which  case  it  will 
be  necessary  to  increase  the  thickness  of  slab  given  in  the  table  to 
the  required  amount. 


&EENFORCED  CONCRETE 


173 


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174  RESISTANCE  OF  MATERIALS 

106.  Dimensions  of  spider.  The  size  and  number  of  rods  in  the 
spider  are  determined  from  the  condition  that  the  total  area  shall 
be  sufficient  to  carry  the  shear,  as  will  now  be  shown. 

It  is  customary  to  place  a  fillet  at  the  top  of  each  column,  as 
shown  in  Fig.  Ill,  the  depth  of  the  fillet  being  not  less  than  the 
thickness  of  the  slab,  and  its  diameter  about  twice  the  diameter  of 
the  column.  The  area  of  concrete  in  shear  at  the  face  of  the  column 
is  then  7r^(2f),  where  d  denotes  the  diameter  of  the  column  and  t 
the  thickness  of  the  slab,  and  at  the  outside  of  the  fillet  is  7r(2  d)t, 
which  is  the  same  amount.  Consequently,  a  fillet  of  these  dimen- 
sions doubles  the  area  of  concrete  in  shear.  The  chief  purpose  of 
the  fillet,  however,  is  to  avoid  the  weakening  effect  of  an  angle 
and  thus  enable  the  slab  to  develop  its  full  strength  at  its  junction 
with  the  column. 

Since  there  are  more  slab  rods  to  carry  the  shear  at  the  outside 
of  the  fillet  than  at  the  face  of  the  column,  the  latter  section  is 
weakest  in  shear.  To  illustrate  the  method  of  dimensioning  for 
shear,  consider  the  same  numerical  problem  as  above ;  namely,  a 
panel  20  x  20  ft.  with  a  total  live  and  dead  load  of  320  lb./ft.2 
The  total  load  on  each  column,  or  the  total  shear,  is  then 
320  X  400=128,000  Ib.  Assuming  that  the  columns  supporting 
the  floor  are  1  ft.  in  diameter,  and  taking  a  cylindrical  section  at 
the  face  of  the  column,  the  total  area  of  concrete  in  shear,  includ- 
ing the  fillet,  will  be  TT  x  12  x  2  x  9.5  =  716  sq.  in.  For  a  working 
stress  in  shear  of  30  lb./in.2,  the  shearing  strength  of  the  concrete 
alone  is  therefore  716  x  30  =  21,480  Ib. 

The  amount  of  metal  in  the  slab  rods  was  determined  previously 
as  0.628  in.2/ft.  Since  the  column  is  1  ft.  in  diameter,  and  since 
each  set  of  rods  is  in  double  shear  and  there  are  four  sets  of 
rods,  the  total  area  of  metal  in  shear  at  the  surface  of  the  column 
is  8  x  0.628  =  5  sq.  in.  Assuming  the  working  stress  in  shear  of 
the  metal  as  10,000  lb./in.2,  the  shearing  strength  developed  by 
the  slab  rods  alone  is  5  x  10,000  =  50,000  Ib.  Since  the  total  load 
is  126,000  Ib.,  there  still  remains  to  be  taken  care  of  126,000  - 
(21,480 +  50,000)- 54,520  Ib.  of  shear. 

To  design  the  spider  to  carry  this  shear,  assume  that  it  is  made 
up  of  8  rods,  as  shown  in  Fig.  111.  Then  the  amount  of  shear  on 


REENFOBCED  CONCRETE  175 

each  rod  will  be  -        —  =  6815  lb.,  and  hence  the  required  area  of 

8 

each  rod  will  be  ^77777:7:  =0.68  sq.  in.,  giving  a  rod  slightly  less 

than  1  in.  in  diameter.  If  the  fillet  under  the  head  is  neglected  in 
computing  the  shearing  strength  of  the  concrete,  a  spider  made  up  of 
eight  1-in.  rods  will  still  give  sufficient  area  to  develop  the  required 
shearing  strength. 

APPLICATIONS 

267.  A  plain  concrete  beam  6  in.  x  6  in.  in  cross  section,  and  with  a  68-in.  span, 
is  supported  at  both  ends  and  loaded  in  the  middle.   The  load  at  failure  is  1008  lb. 
Find  the  maximum  fiber  stress. 

268.  A  concrete  building  block  24  in.  in  length  and  having  an  effective  cross 
section  of  8  in.  x  10  in.  minus  4  in.  x  6  in.  is  tested  by  being  supported  at  points  2  in. 
from  each  end  and  loaded  in  the  middle.  The  load  at  failure  is  found  to  be  5000  lb. 
Find  the  maximum  fiber  stress,  the  height  of  the  block  being  10  in. 

269.  A  reenforced  concrete  beam  10  in.  wide  and  22  in.  deep  has  four  1^-in. 
round  bars  with  centers  2  in.  above  the  lower  face.    The  span  is  16  ft.   The  beam 
is  simply  supported  at  the  ends.   Find  the  safe  load  per  linear  foot  for  a  working 
stress  in  the  concrete  of   500  lb./in.2,  and  also  find  the  tensile  stress  in  the 
reenforcement. 

270.  A  reenf  orced-concrete  beam  of  16  ft.  span  is  12  in.  wide  and  has  to  support 
a  uniform  load  of  1000  lb.  per  linear  foot.    Determine  the  total  depth  and  amount  of 
steel  reenforcement  required,  bars  to  have  centers  2  in.  above  the  lower  face  of  beam. 

271.  A  reenforced  concrete  beam  8  in.  x  10  in.  in  cross  section,  and  15  ft.  long, 
is  reenforced  on  the  tension  side  by  six  i-in.  plain  steel  rounds.    The  steel  has  a 
modulus  of  elasticity  of  30,000,000  lb./in.2  and  the  center  of  the  reenforcement  is 
placed  2  in.  from  the  bottom  of  the  beam.    Assuming  that  Ec  =  3,000,000  lb./in.2, 
and_pc  =  600  lb./in.2,  find  the  position  of  the  neutral  axis  and  the  moment  M. 

272.  For  a  stress  pc  —  2700  lb./in.2  on  the  outer  fiber  of  concrete  in  the  beam 
given  in  problem  271,  find  the  stress  ps  in  the  steel  reenforcement. 

273.  A  concrete  beam  is  10  x  16  in.  in  cross  section  and  20  ft.  long.    It  is  reen- 
forced with  four  5-in.  steel  rods  with  centers  2  in.  above  the  lower  face  of  the 
beam.   The  safe  compressive  strength  of  the  concrete  is  600  lb./in.2,  and  the  steel 
used  has  an  elastic  limit  of  40,000  lb./in.2    What  single  concentrated  load  will  the 
beam  carry  at  its  middle  ?    What  tension  will  be  developed  in  the  steel  ?    What 
shearing  stress  along  the  reenforcement  ? 

274.  Find  what  load,  uniformly  distributed,  the  beam  in  the  preceding  prob- 
lem will  carry  and  find  the  tension  in  the  steel  and  bond  for  this  case. 

275.  A  reenforced  concrete  floor  is  to  carry  a  load  of  200  lb./ft.2  over  panels 
14  ft.  square.    Find  the  required  thickness  of  the  slab  and  the  area  of  the  reenforce- 
ment for  working  stresses  of  500  lb./in.2  in  the  concrete  and  15,000  lb./in.2  in  the 
reenforcement. 

276.  Design  a  floor  panel  14  ft.  square,  to  be  made  of  reenforced  concrete  and 
to  sustain  a  total  uniform  load  of  120  lb./ft.2,  with  a  factor  of  safety  of  4. 


SECTION  XIV 


SIMPLE  STRUCTURES 

107.  Composition  and  resolution  of  forces.  It  will  now  be  neces- 
sary to  recall  some  of  the  results  previously  obtained  concerning 
the  composition  and  resolution  of  forces. 

It  was  shown  in  article  11  that  any  number  of  concurrent  forces 
may  be  combined  by  means  of  a  vector  triangle  or  vector  polygon 
into  a  single  resultant.  Also  that,  conversely,  any  force  may  be 
resolved  into  components  forming  with  the  given  force  a  closed 

triangle  or  polygon. 

In  finding  the  resultant 
of  several  forces  it  is  usually 
more  convenient  to  resolve 
each  of  the  given  forces  into 
components  parallel  to  a  set 
of  rectangular  axes,  then 
take  the  algebraic  sum  of  the 

components  along  each  axis, 

and,  finally,  recombine  these 
into  the  required  resultant. 

Thus,  in  Fig.  115,  if  F^  F^  denote  two  forces  and  R  their  result- 
ant, resolve  F1  into  rectangular  components  x^  y^  and  FZ  into  com- 
ponents x^  yz.  Then,  if  x,  y,  denote  the  components  of  the  resultant 
J?,  we  have 

and,  consequently, 


FIG.  115 


In  article  10  the  moment  of  a  force  with  respect  to  any  point 
was  defined  as  the  product  of  the  force  by  its  perpendicular  distance 
from  the  point  in  question ;  that  is, 


Moment  =  force  X  lever  arm. 

176 


SIMPLE   STRUCTUEES 


177 


It  was  also  proved  that  the  sum  of  the  moments  of  any  number  of 
forces  lying  in  the  same  plane  with  respect  to  a  point  in  this  plane 
is  equal  to  the  moment  of  their  resultant  with  respect  to  this  point. 

It  remains  to  consider  the  case  when  the  system  of  forces  lie  in 
the  same  plane  but  are  not  concurrent,  that  is,  do  not  all  meet  in 
a  point.  This  involves  the  properties  of  a  force  couple,  denned  as 
two  equal  and  opposite  parallel  forces  F,  F,  not  acting  in  the  same 
line  (Fig.  116). 

For  any  couple  F,  F,  let  x  denote  the  distance  of  any  point  0  in 
its  plane  from  the  nearest  force  of  the  couple,  and  d  the  lever  arm 
of  the  couple  (Fig.  116).  Then  the  moment  M  of  the  couple  with 
respect  to  the  point  0  is 


Therefore  the  moment  of  the 
couple  is  constant  and  equal  to 
Fd  with  respect  to  any  point  in 
its  plane.  Moreover,  since  the 
moment  of  the  couple  involves  FlG  116 

only  the  magnitude  of  the  forces 

and  their  distance  apart,  it  is  evident  that  the  couple  can  be  revolved 
through  any  angle  without  altering  its  value.  A  couple  may,  there- 
fore, be  moved  about  anywhere  in  its  plane  without  altering  its 
numerical  value  or  changing  its  effect  in  any  way. 

It  is  also  obvious  that  the  forces  of  a  couple  may  be  altered  in 
amount,  provided  that  the  lever  arm  is  at  the  same  time  changed 
so  as  to  keep  their  product  constant.  Two  or  more  couples  may 
therefore  be  combined  by  first  reducing  them  to  equivalent  couples 
having  the  same  lever  arm  and  then  taking  the  algebraic  sum 
of  the  forces,  thus  giving  a  single  resultant  couple  with  this  same 
lever  arm. 

Now  consider  any  number  of  forces  F^  F^  F8  •  •  •,  lying  in  the  same 
plane  but  not  concurrent.  At  any  arbitrary  point  0(Fig.  117),  intro- 
duce two  forces  F^,  F",  opposite  in  direction,  but  each  equal  in 
amount  to  F^  Since  Ff  and  F"  are  equal  and  opposite  they  will 
not  disturb  the  equilibrium  of  the  system.  But  Fl  and  F"  together 
form  a  couple  of  moment  FI  d^  leaving  the  single  force  F^,  equal 


178  RESISTANCE  OF  MATERIALS 

to  F^  acting  at  0.  Similarly,  each  of  the  other  forces  is  equivalent 
to  a  couple  plus  a  single  force  (equal  and  parallel  to  the  given 
force)  acting  at  0.  The  given  force  system  is  therefore  equivalent 

to  a  system  of  equal  but  concur- 
rent forces  acting  at  0,  and  an 
equal  number  of  couples,  the 
moment  of  each  couple  being 
equal  to  the  moment  of  the  cor- 
responding given  force  with  re- 
spect to  the  point  0. 

This  concurrent  force  system, 
however,  may  now  be  combined 
.  117  into  a  single  resultant  force,  and 

the  couples  also   combined  into 

a  single  resultant  couple,  as  just  explained.  Consequently,  we 
have  the  following  general  theorem: 

Any  system  of  forces  lying  in  the  same  plane  is  equivalent  to  a  single 
force  acting  at  any  assigned  point  in  this  plane  plus  a  couple  whose 
moment  is  equal  to  the  sum  of  the  moments  of  the  given  forces  with 
respect  to  this  point. 

108.  Conditions  of  equilibrium  of  a  system  of  coplanar  forces. 
When  a  body  acted  upon  by  two  or  more  forces  is  at  rest  or  in 
uniform  motion  relative  to  any  system  of  coordinate  axes,  it  is 
said  to  be  in  equilibrium,  and  the  forces  acting  on  it  are  said  to 
equilibrate.  The  conditions  for  equilibrium  are,  therefore,  that  the 
resultant  force  acting  on  the  body  must  be  zero,  and  that  the  result- 
ant moment  or  couple  acting  on  it  must  also  be  zero.  That  is  to 
say,  the  algebraic  sum  of  all  the  forces  acting  on  the  body  must  be 
zero,  and  the  algebraic  sum  of  the  moments  of  these  forces  with 
respect  to  any  point  must  also  be  zero.  Expressed  symbolically  the 
conditions  of  equilibrium  are 


In  general  it  is  convenient  in  applying  these  conditions  to  resolve 
each  force  F  into  rectangular  components  X,  Y,  and  replace  the  single 
condition  ^  F  =  0  by  the  two  independent  conditions  V  X  =  0, 


SIMPLE  STRUCTURES  179 

These  three  conditions,  ^X=  0>  2)  Y==  ^  ^M==  ^  are  obviously 
both  necessary  and  sufficient  to  assure  equilibrium.  For  if  the 
first  two  are  satisfied,  the  system  will  be  in  equilibrium  as  regards 
translation,  and  if  ^?M=  0,  it  will  also  be  in  equilibrium  as 
regards  rotation  ;  and,  furthermore,  it  will  not  be  in  equilibrium  un 
less  all  three  are  satisfied. 

The  conditions  for  equilibrium  of  a  system  of  forces  lying  in  the 
same  plane  may  then  be  reduced  to  the  following  convenient  form : 

1.  For  equilibrium  against  translation, 

I  V  horizontal  components  =  O, 
j  V  vertical  components        =  O. 

2.  For  equilibrium  against  rotation, 

^P  moments  about  any  point  =  O. 

When  a  body  is  acted  on  by  only  three  forces,  lying  in  the  same 
plane,  the  conditions  for  equilibrium  are  that  these  three  forces  shall 
meet  in  a  point,  and  that  one  of  them  shall  be  equal  and  opposite 
to  the  resultant  of  the  other  two. 

109.  Equilibrium  polygon.  As  explained  above,  the  resultant  of 
any  system  of  forces  lying  in  the  same  plane  may  be  found  by  means 
of  a  vector  force  polygon,  the  resultant  being  the  closing  side  of 
the  polygon  formed  on  the  given  system  of  forces  as  adjacent  sides. 
Although  this  construction  gives  the  magnitude  and  direction  of 
the  resultant,  it  does  not  determine  its  position  or  its  line  of  action. 
The  most  convenient  way  to  determine  the  line  of  action  of  the 
resultant  is  to  introduce  into  the  given  system  two  equal  and  oppo- 
site forces  of  arbitrary  amount  and  direction,  such  as  P'  and  P" 
(Fig.  118).  Since  P'  and  P"  balance  one  another,  they  will  not 
affect  the  equilibrium  of  the  given  system.  To  find  the  line  of 
action  of  the  resultant  R,  combine  P'  and  P±  into  a  resultant  Rl 
acting  along  B'A',  parallel  to  the  corresponding  ray  OB  of  the  force 
polygon.  Prolong  A'B'  until  it  intersects  P2  and  then  combine  RI 
and  Pz  into  a  resultant  RZ  acting  along  C'B',  parallel  to  the  corre- 
sponding ray  OC  of  the  force  polygon,  etc.  Proceed  in  this  way 
until  the  last  partial  resultant  R±  is  obtained.  Then  the  resultant 


180 


RESISTANCE  OF  MATERIALS 


of  P'  and  R^  will  give  the  line  of  action,  as  well  as  the  magnitude,  of 
the  resultant  of  the  original  system  P^  P2,  Pz,  Pv  The  closed  figure 
A'B'  C'D'E'F1  obtained  in  this  way  is  called  an  equilibrium  polygon. 


E 


FIG. 118 


For  a  system  of  parallel  forces  the  equilibrium  polygon  is  con- 
structed in  the  same  manner  as  above,  the  only  difference  being 
that  in  this  case  the  force  polygon  becomes  a  straight  line  (Fig.  119). 


FIG. 119 


Since  P'  and  P"  are  entirely  arbitrary  in  both  magnitude  and 
direction,  the  point  0,  called  the  pole,  may  be  chosen  anywhere 
in  the  plane.  Therefore,  in  constructing  an  equilibrium  polygon 


SIMPLE   STRUCTURES  181 

corresponding  to  any  given  system  of  forces,  the  force  polygon 
ABODE  (Figs.  118  and  119)  is  first  drawn,  then  any  convenient 
point  0  is  chosen  and  joined  to  the  vertices  A,  B,  C,  D,  E,  of  the 
force  polygon,  and  finally  the  equilibrium  polygon  is  constructed 
by  drawing  its  sides  parallel  to  the  rays  OA,  OB,  OC,  etc.  of  the 
force  diagram.  Since  the  position  of  the  pole  0  is  entirely  arbitrary, 
there  is  an  infinite  number  of  equilibrium  polygons  corresponding 
to  any  given  set  of  forces.  The  position  and  magnitude  of  the 
resultant  E,  however,  is  independent  of  the  choice  of  the  pole, 
and  will  be  the  same  no  matter  where  0  is  placed. 

For  a  system  of  concurrent  forces  (that  is,  forces  which  all  pass 
through  the  same  point)  the  closing  of  the  force  polygon  is  the 
necessary  and  sufficient  condition  for  equilibrium.  If,  however,  the 
forces  are  not  concurrent,  or  if  they  are  parallel,  this  condition  is 
necessary  but  not  sufficient,  for  in  this  case  the  given  system  of 
forces  may  be  equivalent  to  a  couple,  the  effect  of  which  would  be 
to  produce  rotation.  To  assure  equilibrium  against  rotation,  there- 
fore, it  is  also  necessary  that  the  equilibrium  polygon  shall  close. 

The  graphical  and  analytical  conditions  for  equilibrium  are  then 
as  follows : 

CONDITIONS  OF  EQUILIBRIUM 


Analytical 

Graphical 

Translation 

2>=o 

Force  polygon  closes 

Rotation 

2^=o 

Equilibrium  polygon  closes 

110.  Application  of  equilibrium  polygon  to  determining  reactions. 

One  of  the  principal  applications  of  the  equilibrium  polygon  is  in  de- 
termining the  unknown  reactions  of  a  beam  or  truss.  To  illustrate 
its  use  for  this  purpose,  consider  a  simple  beam  placed  horizontally 
and  bearing  a  number  of  .vertical  loads  P±,  P2,  etc.  (Fig.  120).  To 
determine  the  reactions  E^  and  R^  the  force  diagram  is  first  con- 
structed by  laying  off  the  loads  I±,  7J,  etc.  to  scale  on  a  line  AF, 
choosing  any  convenient  point  0  as  pole  and  drawing  the  rays  OA, 
OB,  etc.  The  equilibrium  polygon  corresponding  to  this  force 
diagram  is  then  constructed,  starting  from  any  point,  say  A',  in  R^. 


182 


RESISTANCE  OF  MATERIALS 


Now  the  closing  side  A'  G'  of  the  equilibrium  polygon  determines 
the  line  of  action  of  the  resultants  P'  and  P"  at  A'  and  G'  respec- 
tively. For  a  simple  beam,  however,  the  reactions  are  vertical. 
Therefore,  in  order  to  find  these  reactions,  each  of  the  forces  P'  and 
P"  must  be  resolved  into  two  components,  one  of  which  shall  be 
vertical.  To  accomplish  this,  suppose  that  a  line  OH  is  drawn  from 
the  pole  0  in  the  force  diagram  parallel  to  the  closing  side  G'A'  of 
the  equilibrium  polygon.  Then  HO  (or  P')  may  be  replaced  by 
its  components  HA  and  A  0,  parallel  to  R^  and  A'B'  respectively ; 


B 


FIG.  120 

and  similarly,  OH  may  be  replaced  by  its  components  FH  and  OF, 
parallel  to  RZ  and  Ff  G'  respectively.  HA  and  FH  are  therefore  the 
required  reactions. 

111.  Equilibrium  polygon  through  two  given  points.  Let  it  be 
required  to  pass  an  equilibrium  polygon  through  two  given  points, 
say  M  and  N  (Fig.  121). 

To  solve  this  problem  a  trial  force  diagram  is  first  drawn  with 
any  arbitrary  point  0  as  pole,  and  the  corresponding  equilibrium 
polygon  M A'B'  C'D'E'  constructed,  starting  from  one  of  the  given 
points,  say  M.  The  reactions  are  then  determined  by  drawing  a 
line  OH  parallel  to  the  closing  side  ME'  of  the  equilibrium  polygon, 
as  explained  in  the  preceding  article. 

The  reactions,  however,  are  independent  of  the  choice  of  the  pole 
in  the  force  diagram,  and,  consequently,  they  must  be  of  amount  AH 
and  HE,  no  matter  where  0  is  placed.  Moreover,  if  the  equilibrium 


SIMPLE   STRUCTURES 


183 


polygon  is  to  pass  through  both  M  and  N,  its  closing  side  must 
coincide  with  the  line  MN,  and  therefore  the  pole  of  the  force  dia- 
gram must  lie  somewhere  on  a  line  through  H  parallel  to  MN. 


FIG.  121 

Let  0'  be  any  point  on  this  line.  Then,  if  a  new  force  diagram  is 
drawn  with  0'  as  pole,  the  corresponding  equilibrium  polygon 
starting  at  M  will  pass  through  N. 

112.  Equilibrium  polygon  through  three  given  points.  Let  it  be 
required  to  pass  an  equilibrium  polygon  through  three  given  points, 
say  M,  N,  and  L  (Fig.  122). 


Fig.  122 


As  in  the  preceding  article,  a  trial  force  diagram  is  first  drawn 
with  any  point  0  as  pole,  and  the  corresponding  equilibrium  polygon 
constructed,  thus  determining  the  reactions  Rl  and  R^  as  HA  and 
EH  respectively. 


184  RESISTANCE  OF  MATERIALS 

Now,  if  the  equilibrium  polygon  is  to  pass  through  N,  the  pole  of 
the  force  diagram  must  lie  somewhere  on  a  line  HK  drawn  through 
H  parallel  to  MN,  as  explained  in  the  preceding  article.  The  next 
step,  therefore,  is  to  determine  the  position  of  the  pole  on  this  line 
HK,  so  that  the  equilibrium  polygon  through  M  and  N  shall  also 
pass  through  L.  This  is  done  by  drawing  a  vertical  LS  through  L 
and  treating  the  points  M  and  L  exactly  as  M  and  N  were  treated. 
Thus  OABCD  is  the  force  diagram  for  this  portion  of  the  original 
figure,  and  MA'B'  C' S  is  the  corresponding  equilibrium  polygon,  the 
reactions  for  this  partial  figure  being  H'A  and  DH'.  If,  then,  the 
equilibrium  polygon  is  to  pass  through  L,  its  closing  side  must  be 
the  line  ML,  and  consequently  the  pole  of  the  force  diagram  must 
lie  on  a  line  H'K'  drawn  through  H'  parallel  to  ML.  The  pole  is 
therefore  completely  determined  as  the  intersection  0'  of  the  lines 
HK  and  H'K'.  If,  then,  a  new  force  diagram  is  drawn  with  0'  as 
pole,  the  corresponding  equilibrium  polygon  starting  from  the  point 
M  will  pass  through  both  the  points  L  and  N. 

Since  there  is  but  one  position  of  the  pole  0',  only  one  equi- 
librium polygon  can  be  drawn  through  three  given  points.  In  other 
words,  an  equilibrium  polygon  is  completely  determined  by  three 
conditions. 

113.  Application  of  equilibrium  polygon  to  calculation  of  stresses. 
Consider  any  structure,  such  as  an  arch  or  arched  rib,  supporting 
a  system  of  vertical  loads,  and  suppose  that  the  force  diagram  and 
equilibrium  polygon  are  drawn  as  shown  in  Fig.  123.  Then  each 
ray  of  the  force  diagram  is  the  resultant  of  all  the  forces  which 
precede  it  and  acts  along  the  segment  of  the  equilibrium  polygon 
parallel  to  this  ray.  For  instance,  OC  is  the  resultant  of  all  the 
forces  on  the  left  of  Pz  and  acts  along  C'D'.  Consequently,  the 
stresses  acting  on  any  section  of  the  structure,  say  mn,  are  the  same 
as  would  result  from  a  single  force  OC  acting  along  C'D'. 

Let  9  denote  the  angle  between  the  segment  C'D'  of  the  equi- 
librium polygon  and  the  tangent  to  the  arch  at  the  point  S.  Then 
the  stresses  acting  on  the  section  mn  at  S  are  due  to  a  tangential 
thrust  of  amount  OC  cos  6  ;  a  shear  at  right  angles  to  this,  of  amount 
OC  sin  6 ;  and  a  moment  of  amount  OC  •  d,  where  d  is  the  perpen- 
dicular distance  of  C'D'  from  S. 


SIMPLE  STRUCTURES 


185 


From  Fig.  123  it  is  evident  that  the  horizontal  component  of  any 
ray  of  the  force  diagram  is  equal  to  the  pole  distance  OH.  There- 
fore, if  OC  is  resolved  into  its  vertical  and  horizontal  components, 
the  moment  of  the  vertical  component  about  S  is  zero,  since  it  passes 
through  this  point ;  and  hence  the  moment  OC  •  d  =  OH  •  z,  where 
z  is  the  vertical  intercept  from  the  equilibrium  polygon  to  the 
center  of  moments  S.  Having  determined 
the  moment  at  any  given  point,  the  stresses 
at  this  point  can  easily  be  calculated. 

/ 

B 


1) 


114.  Relation  of  equilibrium  polygon  to  bending  moment  diagram. 

In  the  preceding  article  it  was  proved  that  the  moment  acting  at 
any  point  of  a  structure  is  equal  to  the  pole  distance  of  the  force 
diagram  multiplied  by  the  vertical  intercept  on  the  equilibrium 
polygon  from  the  center  of  moments.  For  a  system  of  vertical  loads, 
however,  the  pole  distance  is  a  constant.  Consequently,  the  moment 
acting  on  any  section  is  proportional  to  the  vertical  intercept  on  the 
equilibrium  polygon  from  the  center  of  moments.  Therefore,  if  the 
equilibrium  polygon  is  drawn  to  such  a  scale  as  to  make  this  factor 
of  proportionality  equal  to  unity,  the  equilibrium  polygon  will  be 
identical  with  the  bending  moment  diagram  for  the  given  system 
of  loads. 

115.  Structures :    external  forces.    The   external  forces    acting 
upon  any  stationary  structure  must  be  in  equilibrium.   Hence  they 
may  be  found,  in  general,  by  applying  the  conditions  of  equilibrium 
given  in  article  109.   The  conditions  of  equilibrium  may  be  applied 
either  analytically  or  graphically.    The  former  method  has  the  ad- 
vantage of  being  available  under  all  circumstances ;   whereas  the 


186 


RESISTANCE  OF  MATERIALS 


FIG. 124 


latter  method  requires  the  accurate  use  of  instruments,  and  is 
therefore  confined  chiefly  to  office  work.  Both  methods  are  illus- 
trated in  what  follows. 

1.  Analytical  method.  Consider  first  the  analytical  determination 
of  the  external  forces  acting  on  a  simple  structure,  such  as  the 
loaded  jib  crane,  shown  in  Fig.  124.  This  consists  of  a  vertical 

>A  mast  ED,  supported  by  a  collar 
B  and  footstep  (7,  and  carrying 
a  jib  AD,  supported  by  the  guy 
AEF.  The  external  forces  acting 
on  the  crane  are  the  load  W, 
the  counterweight  W^  (including 
hoisting  engine  and  machinery), 
and  the  reactions  at  B  and  C. 
The  reaction  of  the  collar  B  can 
have  no  vertical  component,  as 
the  collar  is  made  a  loose  fit  so 
that  the  crane  may  be  free  to 
swivel.  For  convenience,  the  reaction  of  the  footstep  C  may  be 
replaced  by  its  horizontal  and  vertical  components  H  and  V. 

Applying  the  conditions  of  equilibrium  to  the  structure  as  a 
whole,  we  have,  therefore, 

vertical  forces  =  0,     W  +  ~tt\  -\-  weight  of  crane  —  V=  0, 

horizontal  forces  =0,     H1  +  Jf2  =  0, 

moments  =  0  (taken  about  IT),      Wlz  —  W^  -H  H^c  =  0. 

From  the  first  condition  the  vertical  reaction  of  the  footstep  is 
found  to  be  equal  to  the  entire  weight  of  the  structure  and  its 
loads.  In  applying  the  last  condition,  moments  are  taken  about 
B,  since  one  unknown  Hz  is  thus  eliminated,  leaving  the  resulting 
moment  equation  with  only  one  unknown  H^  The  other  unknown 
H2  is  then  found  from  the  second  condition,  Hz  =  —  Jf^ 

The  moment  of  the  counterweight  W^  should,  when  possible, 

be  made  equal  to  —^i  where  W  is  the  maximum  load  the  crane  is 

designed  to  lift.  The  mast  will  then  never  be  subjected  to  a  bending 
moment  of  more  than  one  half  that  due  to  the  lifted  load  ;  that  is  to 


SIMPLE   STRUCTURES 


187 


say,  the  horizontal  reactions  HI  and  Jf2  will  never  have  more  than  one 
half  the  value  they  would  have  if  the  crane  was  not  counterweighted. 

2.  Graphical  method.  To  illustrate  this  method,  consider  the 
Pratt  truss,  shown  in  Fig.  125.  Assume  the  loads  in  this  case  to 
be  the  weight  of  the  truss 
IF,  a  uniform  load  of 
amount  Tf^,  assumed  for 
present  purposes  to  be 
concentrated  at  its  cen- 
ter of  gravity,  and  two 
concentrated  loads  JJ,  Pr 
Since  the  only  other  ex- 
ternal forces  acting  on  the 
truss  are  the  reactions  R^ 
Rz,  they  must  hold  the 
loads  in  equilibrium,  and 
hence  the  force  polygon 
must  close.  The  force 
polygon,  however,  con- 
sists in  the  present  case 
simply  of  a  straight  line 
12345,  and  therefore 
does  not  suffice  to  deter- 
mine the  values  of  El  and  Rf  For  this  purpose  an  equilibrium 
polygon  must  be  drawn.  Thus,  choose  any  pole  0  on  the  force 
diagram  and  draw  the  rays  0  1,  02,  03,  etc. ;  then  construct  the 
corresponding  equilibrium  polygon  by  starting  from  any  point  a  in 
R  and  drawing  ab  parallel  to  0 1,  from  I  drawing  be  parallel  to 
0  2,  etc.  Having  found  the  closing  side  af  of  the  equilibrium  poly- 
gon, draw  through  0  the  ray  0  6  parallel  to  <//,  thereby  determining 
7^  as  50  and  Rz  as  61. 

If,  for  any  reason,  it  is  desired  to  draw  the  equilibrium  polygon 
through  two  fixed  points,  say  a  and/'  in  the  figure,  the  reactions 
are  first  determined  as  above.  Then  a  line  is  drawn  through  6 
parallel  to  af,  and  a  pole  0'  is  chosen  somewhere  on  this  line.  The 
closing  side  of  the  equilibrium  polygon  will  then  be  parallel  to  0'6 
(or  #/'),  and  hence  if  the  polygon  starts  at  a,  it  must  end  at/'. 


FIG. 125 


188 


RESISTANCE  OF  MATERIALS 


116.  Structures :  joint  reactions.  Since  all  parts  of  a  structure 
at  rest  are  in  equilibrium,  the  conditions  of  equilibrium  may  evi- 
dently be  applied  to  the  forces  acting  upon  any  portion  of  the 
structure.  This  portion  may  be  a  single  joint,  a  single  member  or 
part  of  a  member,  or  it  may  include  several  joints  and  members. 
The  forces  acting  upon  the  part  considered  may  be  partly  external 
forces  and  partly  internal  forces,  or  stresses,  or  they  may  be  wholly 
stresses. 

As  in  finding  external  reactions,  the  conditions  of  equilibrium 
may  be  applied  either  analytically  or  graphically. 


Plan 


FIG.  126 

1.  Analytical  method.  To  illustrate  this  method,  as  applied  to  the 
joints  of  a  structure,  let  it  be  required  to  find  the  stresses  in  the 
members  of  the  shear  legs,  shown  in  Fig.  126. 

Starting  with  the  joint  A,  the  forces  acting  at  this  point  are  the 
weight  W,  the  tension  P  in  the  guy  AC,  and  the  reaction  of  the 
legs  of  the  A  frame.  To  simplify  the  solution  the  latter  may  be 
assumed  for  the  present  equivalent  to  a  single  force  E  acting  along 
the  center  line  A'C  between  the  legs  of  the  A  frame.  The  condi- 
tions of  equilibrium  applied  to  this  joint  are  then 

V  vertical  forces  =  0,     W+  P  cos  (0  +  <£)  -  R  cos  9  =  0, 
V  horizontal  forces  =  0,    P  sin  (0  +  $)  —  R  sin  6  —  0, 
giving  two  simultaneous  equations  for  R  and  P. 


SIMPLE   STKUCTUKES 


189 


Since  R  is  by  assumption  equivalent  to  the  combined  action  of 
the  two  shear  legs,  the  thrust  T  in  each  may  be  found  by  resolving 
forces  along  R.  Thus  T  cos  a  =  1  R,  which  determines  T,  since  R 
has  already  been  found.  I 

Similarly,  the  force  at 
the  bottom  of  the  shear 
legs  tending  to  make 
them  spread  is  T  sin  a. 

At  the  point  C  the 
forces  acting  are  the  up- 
ward pull  V  on  the  an- 
chorage, the  horizontal 
pull  H  011  it,  and  the 
tension  P  in  the  guy. 
Hence,  applying  the  con- 
ditions of  equilibrium, 
we  have 

IT- P  cos  ft  F=Psinft 

2.  G-raphical  method. 
To  illustrate  the  graph- 
ical calculation  of  stresses 
from  joint  reactions,  con- 
sider the  roof  truss 
shown  in  Fig.  127. 

Since  the  loading   in 
this  case  is  symmetrical,  the  reactions  of  the  supports  will  each 
be  equal  to  half  the  weight  on  the  truss. 

The  most  convenient  notation  is  to  letter  the  spaces  between  the 
various  lines  of  the  diagram.  Each  member  of  the  truss  and  each 
external  force  will  then  be  designated  by  the  adjoining  letters  on 
opposite  sides  of  it,  as  the  member  AH,  the  load  BC,  etc. 

Starting  with  the  left  support,  we  have  three  forces  meeting  at 
a  point.  The  magnitude  of  one,  namely  R^  or  AB,  is  known,  and 
the  directions  of  all  three  are  known.  Hence  the  other  two  can  be 
determined  by  means  of  a  triangle  of  forces.  Thus,  if  ab  is  laid  off  to 
scale  to  represent  R^  and  aj,  bj,  are  drawn  from  a  and  b  parallel 


FIG. 127 


190  KESISTAXCE  OF  MATERIALS 

to  AJ  and  BJ,  they  will  represent  the  stresses  in  these  members 
to  the  same  scale  as  that  to  which  R^  was  laid  off  (Fig.  127,  /). 

Proceeding  to  the  next  joint,  BJIC,  we  have  four  forces  meeting 
at  a  point,  one  of  which,  BJ,  has  just  been  determined,  and  another, 
EC,  is  known.  Hence  the  other  two  are  found  by  drawing  a  force 
polygon,  bjic,  giving  the  stresses  in  CI  and  IJ  (Fig.  127,  //). 

Similarly,  passing  to  the  next  joint,  AJIH,  the  stresses  in  AJ  and 
JI  having  been  found,  those  in  Iff  and  AHm&y  be  determined  from 
the  force  polygon  ajih  (Fig.  127,  ///),  and  finally  for  the  joint  If  CD 
the  remaining  stresses  are  determined  from  the  force  polygon  gJticd 
(Fig.  127,  IV). 

Since  each  force  polygon  contains  one  side  of  each  of  the  others, 
by  placing  these  sides  together  they  may  all  be  combined  into 

one  figure,  as  shown  in 

9  Tons      15  Tons  Fig-   127,    F.     Ill    the    pres- 

ent  case  separate  diagrams 
T  ,  ,     .   .    , 

wQTQ  drawn  i or  each  joint 

to  illustrate  the  method. 
In  practice,  however,  but 
one  diagram,  the  combined 
one,  is  drawn,  as  it  affords  a 

F      128  saving  in  time  and  space  and 

produces  a  neater  and  more 

compact  appearance.  Such  a  figure  is  called  a  Maxwell  diagram. 
117.  Structures :  method  of  sections.  If  a  section  is  passed 
through  a  structure,  cutting  not  more  than  two  members  whose 
stresses  are  unknown,  the  single  condition  that  the  force  polygon, 
drawn  for  the  forces  acting  upon  the  portion  of  the  structure  on 
one  side  of  the  section,  must  close,  will  enable  the  stresses  in  these 
members  to  be  found.  Commencing  at  one  end  of  a  structure  and 
passing  a  section  cutting  but  two  members,  the  stresses  in  these 
can  thus  be  determined.  Then,  passing  a  section  cutting  three 
members,  one  of  which  has  already  been  treated,  the  stresses  in  the 
other  two  can  be  found,  etc.  Thus,  by  means  of  successive  sections, 
all  of  the  stresses  can  be  determined  by  simple  force  polygons. 

1.  Analytical  method.  To  illustrate  the  analytical  application  of 
this  method,  consider  a  Warren  truss  used  as  a  deck  bridge,  as 


SIMPLE   STRUCTURES  191 

shown  in  Fig.  128.  Let  the  depth  of  truss  and  panel  length  be 
each  15  ft.,  and  the  loads  carried  a,t  the  joints  of  the  upper  chord 
be  7,  10,  9,  and  15  tons  respectively.  The  reactions  at  B  and  J 
are  found  by  taking  moments  about  J  and  B  to  be  17|  tons  and 
23|  tons  respectively. 

Since  this  form  of  truss  has  parallel  chords  and  a  single  web 
system,  it  is  not  necessary  to  begin  at  any  particular  point,  but  a 
section  may  be  taken  anywhere,  provided  it  cuts  both  chords  and 
a  single  web  member.  Taking  any 
section  xy,  and  considering  only  the 
portion  of  the  structure  on  one  side  of 
the  section,  the  external  forces  acting 
on  this  portion  will  be  in  equilibrium 
with  the  stresses  P,  Q,  R,  in  the  mem- 
bers cut  (Fig.  129).  Since  Q  is  the 
only  stress  having  a  vertical  compo- 
nent, it  must  equilibrate  the  external 
forces  at  B  and  C.  That  is  to  say, 
from  the  condition  of  equilibrium 

^  vertical  forces  =  0,  we  have  Q  sin  03°  20'  =  17|—  7  ;  whence 
Q  =  11.854  tons  and  is  compressive. 

To  find  P  take  moments  about  D.  Then  since  Q  and  R  botli 
pass  through  D,  their  moments  about  this  point  are  zero;  therefore 

P  x  15  =  17-f  x  15  -  7  x  7.5 ; 

whence  P  =  14i  tons.  By  observing  the  signs  of  the  moments  of 
the  external  forces  at  B  and  C  about  I),  P  is  found  to  act  in  the 
direction  shown  by  the  arrow,  that  is,  in  compression. 

Similarly,  to  find  the  stress  R  in  J>F,  take  the  section  vy  just  to 
the  left  of  E,  then  take  moments  about  E.  Since  P  and  Q  pass 
through  E,  their  moments  about  this  point  are  zero,  and  hence 

.flxl5=17f  x  22.5  -7x15; 

whence  R  =  19.44  tons. 

Since  the  loads  are  vertical,  R  might  also  have  been  found 
from  P  and  Q  by  the  condition  2\  horizontal  forces  =  0  ;  that  is, 
P  +  Q  cos  63°  26'  =  R  ;  whence  R  =  19.427  tons. 


192 


RESISTANCE  OF  MATERIALS 


2.  Graphical  method.  Before  proceeding  with  the  explanation  of 
the  graphical  method  it  will  be  necessary  to  show  how  the  moment 
of  any  number  of  forces  with  respect  to  a  given  point  may  be 
obtained  from  the  equilibrium  polygon. 

Let  PV  P2,  ^,  P»  denote  any  set  of  forces  and  B  the  given  point 
about  which  their  moment  is  required  (Fig.  130).  First  draw  the 
force  polygon  for  these  forces,  choose  any  pole  0,  and  construct 

the  corresponding  equilib- 
rium polygon  abcde.    Now 
in  the  force  diagram,  drop 
a  perpendicular  oh  from  the 
pole  0  on  the  resultant  R. 
This  is  called  the  pole  dis,- 
tance  of  R  and  will  be  denoted  by  H. 
Also,  in  the  equilibrium  diagram  draw 
through  the  given  point  B  a  line  par- 
allel to  R,  making  the   intercept  xy 
on   the    equilibrium    polygon.     Then 
the  triangle  OAE  in  the  force  diagram 
is  similar  to  the  triangle  xey  in  the 
equilibrium  diagram,  and  hence 

r  :  xy  =  H  :  AE, 


D 


or 


Rr  =  H  x  xy. 

But  Rr  is  the  moment  of  the  result- 
ant R  about  B  and  is  equal  to  the 
sum  of  the  moments  of  all  the  given  forces  about  this  point.  The 
following  moment  theorem  may  therefore  be  stated : 

The  moment  of  any  system  of  forces  about  a  given  point  is  equal  to 
the  pole  distance  of  their  resultant  multiplied  by  the  intercept  made 
by  the  equilibrium  polygon  on  a  line  drawn  through  the  given  point 
parallel  to  the  resultant. 

The  moment  of  a  part  of  the  given  set  of  forces  about  any  point 
may  also  be  found  by  this  theorem.  For  example,  let  it  be  required 
to  find  the  moment  of  P^  and  Pz  about  B.  The  resultant  of  Pv  P^,  is 
given  in  amount  by  AC  and  acts  through  the  point/,  as  shown. 
Hence  draw  through  B  a  line  parallel  to  this  partial  resultant, 


SIMPLE   STRUCTURES 


193 


making  the  intercept  mn  on  the  equilibrium  polygon.    Then,  since 
the  triangles  fmn  and  OA  C  are  similar,  we  have 

r' :  mn  =  H' :  E', 

or  E'r'  =  H'  x  mn, 

which  is  the  expression  required  by  the  theorem. 

For  a  system  of  parallel  forces  the  pole  distance  H  is  constant 
and  hence  the  equilibrium  polygon  is  similar  to  the  moment 
diagram  for  the  forces  on 
either  side  of  any  given  point. 
Therefore  the  moment  of  all 
the  forces  on  one  side  of  a 
given  point,  taken  with  re- 
spect to  this  point,  is  equal 
to  the  constant  pole  distance 
H  multiplied  by  the  intercept 
made  by  the  equilibrium  poly- 
gon on  a  vertical  through  the 
point  in  question. 

To  apply  this  method  to  the 
roof  truss  shown  in  Fig.  131, 
for  example,  draw  the  force 
polygon  and  the  correspond- 
ing equilibrium  polygon,  as 
shown  in  the  figure.  Now  take 
any  section  of  the  truss,  such 
as  xy  in  the  figure,  and  take 
moments  of  the  stresses  in  the 
members  cut  about  one  of 
the  joints,  say  B.  Then  the  condition  of  equilibrium 

V  moments  about  B  =  0 
may  be  written 

Moment  of  stress  in  AF-\-^ moments  of  P^,  P2, 
But  by  the  above  theorem 

^moments  of  J\,  P2,  A\,  about  B  =  bbr  x  Oh. 

W  X  Oh 


FIG. 131 


Hence 


Stress  in  AF '  = 


194 


RESISTANCE  OF  MATERIALS 


Similarly,  by  taking  moments  about  A  the  stress  in  BF  is  found  to  be 

aa'  x  Oh 


Stress  in  BF  = 


AT 


and  the  stress  in  J5(7,  with  center  of  moments  at  F,  is 

ccf  X  Oh 
Stress  in  BC— . 


By  observing  the  signs  of  the  moments  the  stresses  in  AB,  BC, 
and  BF  are  found  to  be  compressive  and  that  in  AF  tensile. 

In  the  present  case,  from  symmetry,  the  stresses  in  the  remaining 
members  of  the  truss  are  the  same  as  in  those  already  found.  For 
unsymmetrical  loading  it  would  be  necessary  to  apply  the  above 
method  to  each  individual  member. 


APPLICATIONS 

277.  Two  equal  weights  of  50  Ib.  each  are  joined  by  a  cord  which  passes  over 
two  pulleys  in  the  same  horizontal  line,  distant  12  ft.  between  centers.    A  weight 
of  5  Ib.  is  attached  to  the  string  midway  between  the  pulleys.    Find  the  sag. 

278.  The  rule  used  by  the  makers  of  cableways  for  finding  the  stress  in  the 


cable  is  to  calculate  a  factor  = 


one  half  the  span 


and  multiply  the  load,  assumed 


/- — \ 


twice  the  sag 
to  be  at  the  middle,  by  this  factor.    Show  how  this  formula  is  obtained. 

279.  It  is  usual  to 
allow  a  sag  in  a  cable 
equal  to  one  twen- 
tieth  of   the    span. 
What  does  the  nu- 
merical factor  in  the 
preceding    problem 
become  in  this  case, 
and   how   does   the 
tension  in  the  cable 
compare    with    the 
load? 

280.  Find     the 
relation  between  F 
and  fP  and  the  total 

pull  on  the  upper  support  in  the  systems  of  pulleys  shown  in  Fig.  132. 

281.  In  a  Weston  differential  pulley  two  sheaves,  of  radii  a  and  6,  are  fastened 
together,  and  by  means  of  a  continuous  cord  passing  around  both  and  also  around 
a  movable  pulley,  support  a  weight  W.  Find  the  relation  between  F  and  TF,  neg- 
lecting friction  (Fig.  133). 


SIMPLE   STRUCTURES 


195 


282.  In  a  Weston  differential  pulley  the  diameters  of  the  sheaves  in  the  upper 
block  are  8  in.  and  9  in.   Find  the  theoretical  advantage. 

283.  In  the  differential  axle  shown  in  Fig.  134  the  rope  is  wound  in  opposite 
directions  around  the  two  axles  so  that  it  unwinds  from  one  and  winds  up  on  the 
other  at  the  same  time.    Find  its  mechanical  advantage  if  the  radius  of  the  large 
drum  is  #,  of  the  small  drum  is  r,  and  of  the  crank  is  c. 

284.  A  differential  screw  consists  of  two  screws,  one  inside  the  other.    The 
outer  screw  works  through  a  fixed  block,  and  is  turned  by  means  of  a  lever.    This 
screw  is  cored  out  and  tapped  for  a  smaller  screw  of  less  pitch  which  works  through 
another  block,  free  to  move  along  the  axis  of  the  screw,  but  prevented  from  rotat- 
ing.  Find   the  mechanical  advantage  of  such  a  differential  screw  if  the  lever 
arm    is   3  ft.  long,   the    outer    screw 

has    8  threads  to   the   inch,   and  the 
inner  screw  has  10  threads  to  the  inch 


FIG. 135 


285.  The  bed  of  a  straight  river  makes  an  angle  a  with  the  horizontal.  Taking 
a  cross  section  perpendicular  to  the  course  of  the  river,  the  sides  of  the  valley  are 
inclined  at  an  angle  £  to  the  horizontal.    Find  the  angle  which  the  tributaries  of 
the  river  make  with  it. 

286.  Find  the  least  horizontaHorce  necessary  to  pull  a  wheel  30  in.  in  diameter, 
carrying  a  load  of  500  lb.,  over  an  obstacle  4  in.  high. 

287.  A  steelyard  weighs  61b.  and  has  its  center  of  gravity  in  the  short  arm  at 
a  distance  of  1  in.  from  the  fulcrum,  and  the  center  of  suspension  is  3  in.  from  the 
fulcrum.   The  movable  weight  weighs  4  lb.    Find  the  zero  graduation,  and  the  dis- 
tance between  successive  pound  graduations. 

288.  A  ladder  50  ft.  long,  weighing  75  lb.,  rests  with  its  upper  end  against  a 
smooth  vertical  wall  and  its  lower  end  on  rough  horizontal  ground.    Find  the 
reactions  of  the  supports  when  the  ladder  is  inclined  20°  to  the  vertical. 

289.  A  circular,  three-legged  table,  4  ft.  in  diameter,  weighs  50  lb.  and  carries 
a  load  of  100  lb.  10  inches  from  the  center  and  in  a  line  joining  the  center  and  one 
leg.  Find  the  pressure  between  each  foot  and  the  floor.   Find  also  the  smallest  load 
which  when  hung  from  the  edge  of  the  table  will  cause  it  to  tip  over. 

290.  The  average  turning  moment  exerted  on  the  handle  of  a  screw  driver  is 
120  in.-lb.  The  screw  has  a  square  slot,  but  the  point  of  the  screw  driver  is  beveled 
to  an  angle  of  10°  (Fig.  135).  If  the  point  of  the  screw  driver  is  1  in.  wide,  find  the 
vertical  force  tending  to  raise  the  screw  driver  out  of  the  slot. 


196 


RESISTANCE  OF  MATERIALS 


291.  Three  smooth  cylindrical  water  mains,  each  weighing  500  lb.,  are  placed  in 
a  wagon  box,  two  of  them  just  filling  the  box  from  side  to  side  and  the  third  being 
placed  on  top  of  these  two.    Find  the  pressure  between  the  pipes  and  also  against 
the  bottom  and  sides  of  the  wagon. 

292.  An  engine  is  part  way  across  a  bridge,  the  weights  and  distances  being  as 
shown  in  Fig.  136.   Find  the  reactions  of  the  abutments. 


r 64-0 

FIG. 136 


FIG.  137 


293.  In  the  letter  scales  shown  in  Fig.  137  the  length  of  the  parallel  links  is 
3  in.,  and  the  distance  of  the  center  of  gravity  of  the  moving  parts  below  the  pivot 
0  is  2  in.    If  the  radius  of  the  scale  is  8  in.  and  the  weight  of  the  moving  parts  is 
12  oz.,  find  the  distance  between  successive  ounce 

graduations  on  the  scale. 

294.  A  scale  is  arranged  as  shown  in  Fig.  138. 
Determine  the  relation  between  the  load  and  the 
weight  P.    (Quintenz  scales,  Strassburg,  1821.) 


FIG. 138 


FIG.  139 


Solution.    With  the  given  dimensions  we  have,  by  the  principle  of  moments, 


and 
whence 


Pa  = 


Pa  =  Wb  -  WX(b  --  —  cV 
l\        e  +  d  I 


SIMPLE   STRUCTURES 


197 


Since  this  is  independent  of  x,  the  position  of  the  load  on  the  platform  does  not 
affect  the  result. 


Let  the  dimensions  be  so  proportioned  that  -  — 
W 


e  +  d 


and  also  a  =  10  6.  Then 


P  =  —  A  scale  so  arranged  is  called  a  decimal  scale. 


295.  In  the  toggle-joint  press  shown  in  Fig.  139,  the  length  of  the  hand  lever  is 
=  3£  ft.,  and  12  =  4  in.    If  the  pull  P  =  100  lb.,  find  the  pressure  between  the 

jaws  of  the  press  when  the  toggle  is 
inclined  at  10°  to  the  vertical. 

296.  In  the  crab  hook  shown  in  Fig. 
140,  assume  that  the  load  Q  =  300  lb. 
and  the  coefficient  of  friction  /*  =  .5, 
and  determine  the  kind  and  amount 
of  strain  in  the  members  ED,  AD,  and 
A  13  for  a  =  30°,  /3  =  90°,  a  =  6  in., 
6=3  in.,  and  /  =  18  in.  Show  also 
that  in  order  to  hold  the  weight  with- 
out slipping,  the  condition  which  must 
be  satisfied  is 


2  a  sin  a 


FIG.  140 


297.  A    steam    cylinder    is    20  in. 
in   diameter,    the    steam  pressure    is 

150  lb./in.2,  the  crank  is  18  in.  long,  and  the  connecting  rod  is  5  cranks  long. 
Find  the  stress  in  the  connecting  rod,  pressure  on  cross-head  guides,  and  tangen- 
tial pressure  on  crank  pin  when  the  crank  makes  an  angle  of  45°  with  the  horizontal 
on  the  ff  in  end  "  of  the  stroke.  Find  also  the  maximum  tangential  pressure  on  the 
crank  pin. 

298.  Calculate  the  stresses  in  all 
the  members  of  the  dockyard  crane 
shown  in  Fig.  141  when  carrying  a 
load  of  40  tons. 

299.  Calculate    analytically    the 
stresses  in  the  members  of  the  jib 
crane  shown  in  Fig.  142  when  lift- 
ing a  load  of  28  tons.,  the  dimen- 
sions being  as  given  in  the  figure.  FIG.  141 

300.  In    the    locomotive    crane 

shown  in  Fig.  143,  calculate  the  stresses  in  boom,  mast,  back  stays,  hoisting  line, 
and  boom  line  when  the  boom  is  in  its  lowest  position,  the  dimensions  for  this 
case  being  as  given  in  the  figure. 

301.  Calculate  the  maximum  stresses  in  all  the  members  of  the  stiff -leg  derrick 
shown  in  Fig.  144  when  lifting  a  load  of  10  tons,  the  dimensions  being  as  follows : 
mast  =  25  ft.,  boom  =  38  ft.,  each  leg  =  40  ft.   It  is  customary  to  load  the  sills  with 
stone  to  give  the  necessary  stability..  Find  the  amount  of  stone  required  on  each 
sill  when  lifting  the  10-ton  load. 


FIG.  142 


FIG. 143 


FIG. 144 


198 


200 


RESISTANCE  OF  MATERIALS 


302.  The  testing  machine  shown  in  Fig.  145  is  designed  for  a  maximum  load  on 
the  platform  of  100,000  Ib.  The  dimensions  of  the  various  levers  are  as  shown  in 
Fig.  146,  the  lever  C  being  V-shaped,  with  the  lever  D  hung  inside.  Neglecting  the 


Extreme  position 
of  Weight  - 


Plan,  levers  0  and  D 


FIG.  146 


weights  of  the  arms,  compute  the  weight  of  the  slider  when  in  its  extreme  posi- 
tion required  to  balance  a  load  of  100,000  Ib.  on  the  platform. 

303.  Determine  analytically  the  stresses  in  the  members  CD,  DE,  and  EF  of 
the  curved-chord  Pratt  truss  shown  in  Fig.  147,  assuming  the  load  at  each  panel 

point  to  be  50,000  Ib. 

/  304.  The    roof   truss    shown   in 

Fig.  148  is  anchored  at  one  end  A, 
and  rests  on  rollers  at  the  other 
end  B.  The  span  Z  =  80  ft.,  rise 
h  =  30  ft.,  distance  between  trusses 
b  =  18  ft.  The  weight  of  the  truss 
is  given  approximately  by  the  for- 
mula W  =  g1?  bl'2 ',  the  wind  load,  as- 
sumed to  be  from  the  left,  is  taken 
as  451b./ft.2  of  roof  surface,  and 
the  snow  load  is  30  lb./ft.2  of  hori- 
zontal projection.  Calculate  analyt- 
ically the  reactions  of  the  supports 
due  to  all  loads  acting  on  the  truss. 

305.  In  the    saw-tooth   type    of 
roof  truss  shown  in  Fig.  149,  deter- 
mine analytically  the  stress  in  FH. 

306.  In  the  Pratt  truss  shown  in 
Fig.  150,  the  dimensions  and  loads 

are  as  follows:  span  =  150ft.,  height  =  30ft.,  number  of  panels  =  6.  The  dead 
load  per  linear  foot  in  pounds  for  single-track  bridge  of  this  type  is  given  by  the 
formula  w  =  5 1  +  350,  where  I  denotes  the  span  in  feet ;  the  weight  of  single  track 


FIG. 147 


SIMPLE  STRUCTURES 


201 


may  be  taken  as  400  Ib.  per  linear  foot,  and  live  load  as  3500  Ib.  per  linear  foot. 
Calculate  analytically  the  stresses  in  all  the  members. 

NOTE.   Each  truss  carries  one  half  the  total  load.   In  the  present  case,  therefore,  the 


total  load  per  linear  foot  per  truss  is 


1100  +  400  +  3500 

2 


=  25001b. 


307.  In  the  saw-tooth  type  of  roof  truss  shown  in  Fig.  149,  obtain  graphically 
the  stresses  in  all  the  members,  the  dimensions  being  as  follows  :  span  =  25  ft.,  dis- 
tance apart  of  trusses  =  15  ft.,  and  pitch  of  roof  =  §,  making  the  inclination  of  the 
longer  leg  to  the  horizontal  =  21°48/. 


FIG. 149 


As  the  span  is  short  and  the  roof  comparatively  flat,  it  is  sufficiently  accurate 
to  assume  that  the  combined  action  of  wind  and  snow  is  equivalent  to  a  uniform 
vertical  load,  which  in  the  present  case  may  be  assumed  as  25  lb./ft.2  of  roof.  The 
weight  of  this  type  of  truss  will  be  taken  as  1.5  lb./ft.2  of  roof,  and  the  weight  of 
roof  covering  as  7  lb./ft.2  of  roof.  As  the  top-chord  panel  length  is  8  ft.,  each  panel 
load  will  be  8  x  15  x  33.5  =  4020  Ib.  A 

308.  Analyze  graphically  for  both 
dead  and  snow  loads  the  French  type 
of  roof  truss  shown  in  Fig.  151,  the 
dimensions  being  as  follows :  span 


D 


F 


G 


E 


H 


K 


^M        N        O        P        Q        R    f 
4          I         III          \R 
mL J 


I  =  100  ft.,  rise   h  =  30  ft.,   d  =  5  ft., 
and  distance  apart  of  trusses  b  =  20  ft. 

The  weight  of  truss  in  pounds  for  FIG.  150 

this  type   is    given   by  the    formula 

W  =  5*4  W2,  where  b  and  I  are  expressed  in  feet.  The  weight  of  the  roof  covering 
may  be  assumed  as  15  lb./ft.2  of  roof  surface,  and  the  snow  load  as  20  lb./ft.2  of 
horizontal  projection. 

First  calculate  the  dead  load  carried  at  each  joint,  due  to  weight  of  truss  and 
roof  covering,  and  draw  the  diagram  for  this  system  of  loads.    The  diagram  for 


202 


RESISTANCE  OF  MATERIALS 


snow  load  will  be  a  similar  figure,  and  the  stresses  in  the  two  cases  will  be  propor- 
tional to  the  corresponding  loads.  Hence  the  snow-load  stresses  may  be  obtained 
by  multiplying  each  dead-load  stress  by  a  constant  factor  equal  to  the  ratio  of 
the  loads. 


Graphical  analysis  of  French  truss 
G 


FIG. 161 

In  drawing  the  diagram  start  at  one  abutment,  say  the  left,  and  take  the  joints 
in  order,  thus  determining  the  stresses  in  0-4,  AM,  AB,  PB,  BC,  and  CM.  At  the 
middle  of  the  rafter,  where  the  load  PQ  is  applied,  there  will  be  three  unknowns, 
and  since  these  cannot  all  be  determined  simultaneously,  one  of  the  three  must  be  ob- 
tained by  some  other  means  before  the  construction  can  proceed.  For  this  purpose, 


SIMPLE   STRUCTURES 


203 


proceed  to  the  load  QE  and  determine  the  stress  in  EF  by  the  auxiliary  construc- 
tion shown  by  the  dotted  lines  in  the  diagram.  Then  determine  the  stress  in  ED 
by  the  same  auxiliary  construction.  Having  found  the  stress  in  ED,  we  may  then 
go  back  to  the  load  PQ  and  complete  the  diagram. 


Graphical  analysis  of 
Howe  truss 


Wind  load  stress  diagram, 
wind  from  left 


Dead  load  stress  diagram 


FIG.  152 


309.  Analyze  graphically  for  both  dead  and  wind  loads  the  Howe  roof  truss 
shown  in  Fig.  152  for  a  span  of  50  ft.  and  of  one-half  pitch,  that  is,  with  a  rise 
of  ^  =  12.5  ft.  The  trusses  are  spaced  16  ft.  apart ;  the  weight  of  each  truss  may 
be  taken  as  2.5  lb./ft.2  of  horizontal  area,  the  roof  covering  as  10  lb./ft.2  of  roof, 
and  the  snow  load  as  20  lb./ft.2  of  roof.  The  wind  load,  based  on  a  pressure  of 


204  RESISTANCE  OF  MATERIALS 

30  lb./ft.2  of  vertical  projection,  gives  for  a  roof  of  one-half  pitch  an  equivalent 
load  of  22.4  lb./ft.2  of  roof  surface. 

Assume  the  left  end  of  the  truss  to  be  on  rollers  and  the  right  end  fixed.  The 
total  horizontal  thrust  due  to  wind  load  is  then  carried  by  the  right  abutment. 

In  drawing  the  wind-load  diagram,  first  calculate  the  reactions  R^  and  R2  by  the 
method  of  moments.  Having  thus  determined  the  point  M,  the  remainder  of  the 
diagram  is  easily  drawn. 

310.  Solve  problem  304  graphically. 


ANSWERS  TO  PROBLEMS 


The  following  list  comprises  answers  to  about  two  thirds  of  the  problems  given 
at  the  close  of  each  section  as  applications  of  the  text.  This  is  ample  to  enable 
the  student  to  verify  the  correctness  of  his  numerical  applications,  while  enough 
remain  unanswered  to  cultivate  self-reliance  and  independence  of  thought. 


1, 
2. 

7. 

8. 

9. 
10. 
11. 
12. 
13. 
14. 
15. 
16. 
17. 
18. 
19. 
20. 
21. 
22. 
23. 
24. 
25. 
26. 
27. 
28. 
29. 
30. 


5.  3.1  lb./in.2 

6.  .002122;  12. 73  in. 


Answer  given.  3.  Answer  given. 

Answer  given.  4.  17. 7  lb./in.2 

16,500,000  lb./in.2  approximately.          31.  t  = 

.0044  in. approximately^  =.00002037.  32.  Copper,  5.73cm.;  alloy,  4.27cm. 


34.  Brass,  791  Ib.  ;  steel,  409  Ib. 

35.  Eight  £-in.  bolts. 

36.  2  in. 
crushing  _  3 
shearing      TT 

38.  1.7  approximately. 

39.  181. 

40.  f-in.  steel-wire  rope. 

54.  95.45  in.4 

55.  441  in.4 

56.  337  in.4 

57.  Ill  in.4 

58.  1^  in.  above  base. 

59.  83^  in.4 

60.  3.36  in.  from  top. 

61.  171  in. 

62.  576|  ;  364£. 

63.  3.28  in. 

64.  51  in.4 

68.  Weight  equal  to  that  of  table. 

69.  First  support,  If  ft.  from  150  Ib. 

weight. 

77.  Maximum  moment  =  1920  ft.-lb. ;  maximum  shear  =  480  Ib. 

78.  Maximum  moment  =  4225  ft.-lb. ;  maximum  shear  =  950  Ib. 

79.  Maximum  moment  =  4666|  ft.-lb. ;  maximum  shear  =  833^  Ib. 

80.  Maximum  moment  =  2200  ft.-lb. ;  maximum  shear  =  375  Ib. 

81.  Maximum  moment  =  750  ft.-lb. ;  maximum  shear  =  200  Ib. 

82.  Maximum  moment  =  3333^  ft.-lb. ;  maximum  shear  =  8000  Ib. 

83.  Maximum  moment  =  11,850  ft.-lb. ;  maximum  shear  =  4100  Ib. 

93.  Maximum  moment  =  7400  ft.-lb.  ;  maximum  shear  =  1350  Ib. 

94.  Maximum  moment  =  24,000  ft.-lb. ;  maximum  shear  =  4000  Ib. 

95.  Maximum  moment  =  1250  ft.-lb. ;  maximum  shear  =  500  Ib. 

205 


27,708  Ib. 

.000307  in. 

.0005076  in. 

s  =  .0018. 

5. 4  in. 

.0000104. 

1005  Ib. 

.245  in.  square. 

5  in.  for  p  =  16,000  lb./in.2 

113  tons  at  failure. 

20.0086  ft. 

2363  lb./in.2 

.13  in. 

10,980  lb./in.2 

75,280  lb./in.2  for  root  area. 

10|. 

5.8. 

17,440  Ib. 

1,645,300  lb./in.2 

100  +  554  Ib. 

^-in.  pin. 

2036  Ib.  tension  in  each. 


206 


RESISTANCE  OF  MATERIALS 


109. 

110, 

ill, 


35,280  ft.-lb. 
869,500  ft.-lb. 
7  to  4. 


H2.   ±L  ;   — 


113, 
128. 
129. 

130. 
131. 
136. 
166. 
167. 
168. 
169. 
170. 

171. 

173, 
174, 
175. 


205. 
206, 
207. 
222. 
223. 
224. 
225. 
226. 
228. 

229. 
230. 
231. 
232. 
233. 
234. 
235. 


241. 


6       24      32 
Sin.  I,  20|lb. 

18.4  including  weight  of  beam. 
Tension  =  2480  +  500  lb./in.2 
Compression  =  4100  —  500  lb./in.2 


114.  15-in.  channels,  33  Ib. 

115.  1  to  12. 

116.  Section  modulus =24. 

117.  9  in.  I,  21  Ib. 

118.  9  in.  I,  25  Ib. 

119.  6    in. 


51  in. 
16.3ft. 
.192  in. 
350  tons-. 
9  1  in.  square. 
5.82  in. 


123.  830in.4; 
2767  Ib./ft. 

124.  3967  Ib. 

125.  4i  x  f  in. 

126.  28.1  Ib. 

127.  Stress  =  51 70  lb./in.2 

137.  10111b./ft.2 

138.  12.68  ft.  apart. 

139.  .0373  in. 

156,  Moment  at  wall  =  1800 ft.-lb.; 

D  maximum  =  .0776  in.  at  5.06  ft. 
from  free  end. 


176,  4-in.  channels,  5  Jib.;    185.  20  in.  I,  75  Ib. 


Rankine,  616  tons  ; 
Johnson,  627  tons. 
Rankine,  268  tons  ; 
Johnson,  267  tons. 
132  tons. 
15. 

2  1  in.  square. 
d  =  4.25  in.  for  stiffness  ; 
.d  =  6.6  in.  for  strength. 
Minimum  speed,  73  R.  P.  M. 
2500  lb./in.2 
Shaft,  3  in.  diameter. 
.13  in. 

5911  lb./in.2 
97J. 

2940  lb./in.2 
3333  lb./in.2 
Side,  ^j  in.; 


plates,  6  x  \  in. 
177.  Top  plate,  10  x  £i 
side  plates,  8^  x  ^  in 
angles,  3  x  3  x  |  in 

180.  3.4  in.;  .9  in. 

181.  1=  17.  Id. 

182.  Diameters,  10^  and 

71  in. 

183.  About  3  in. 

184.  5  +  . 


bottom,  ^9¥  in. 
6528  lb./in.2 

5880  lb./in.2 

4ft. 

lin. 

2  in. 

4. 

3677  lb./in.2; 

98.9  tons ; 

13,550  lb./in.2; 

494.6in.-tons. 

1.25  in. 


221.  ^ 
242.  2344  lb./in.2 

244.  139  lb./in.2 

245.  .28  in. 

246.  17001b./ft.2 

247.  t  =  .108r. 

248.  Flat,  t  =  3. 16 in.; 
hemispherical,  £  =  .7  in. 

250.  1.28  in. 

261.  70.8%;  52.4%. 

262.  J-in.  plates, 
|-in.  rivets, 
3-in.  pitch ; 

e  =  75  to  78  %. 

263.  ^-in  plates ; 
1-in.  rivets, 
5.2-in.  pitch. 

264.  82.5  lb./in.2 

267.  476  lb./in.2 

268.  450  lb./in2. 


190.  43.24  in.-lb. 

192.  3.68  in. 

193.  5.63  in. 

194.  7. 114  in. 

195.  32°  28'. 

196.  2|  H.  P. 

197.  4.465  in. 

200.  4484  H.  P. 

201.  9.2  in. 

202.  32. 

208.  p  =  4235  lb./in.2; 
q  =  4283  lb./in.2 

209.  .E  =  30,346,935  lb./in.2; 
G  =  12,158,485  lb./in.2 

in. 

269.  7681b./ft. 
277.  .3ft. 
279.  5  times  load. 
W 


280.  F=—    " 


_  W  / 
=  2\ 


a,  -  b 


281.  F 

282.  1  to  18. 

283.F  =  E/£r 


-0: 


2  \    c 

284.  W=  90478  J?,  neglect- 
ing friction. 


sin  a 
286.  463.2  Ib. 

288.  Upper  end,  13.6  Ib.; 
lower  end,  76.2  Ib. 

289.  22flb.,63flb.,63|lb.; 
50  Ib. 

290.  82  Ib. 


ANSWERS  207 

291.  288.7  Ib.  between  pipes  ;  298.  AK  =  86  tons,  AB  =  76.3  tons, 
144.35  Ib.  against  sides.  EF  =  121.6  tons,  ED  =  82.9  tons. 

292.  Left,  174,110  Ib.;  299.  AD  =  99.2  tons,  BC  =  95.6  tons, 
right,  235,992  Ib.  AB  =  87. 7  tons,  A  C  =  85.5  tons, 

295.  6.83  tons.  DE  =  49.5  tons,  FC  =  117.0  tons, 

297.  Connecting  rod,  47,580  Ib. ;  DF  =  51.7  tons. 

cross  head,  6730  Ib.  303.   CD  =  296,500  Ib.,  DE  =  12,842  Ib., 

Maximum  on  crank  pin  48,030  Ib.  EF  =  288,500  Ib. 

304.  BH  =  49,400  Ib.,  BE  =  33,450  Ib.,  HE  =  31,090  Ib.,    HC  =  33,500  Ib., 
CE  =  24,657  Ib.,  CK  =  41,094  Ib.,  KE  =  10,000  Ib.,  KA  =  51,094  Ib., 

AE=  16,574  Ib. 

305.  1390  Ib. 

306.  AB  =  203,100  Ib.,  AD  =  208,300  Ib.,  AG  =  234,375  Ib., 
BM  =  CN  =  130,200  Ib.,  #0  =  208,300  Ib.,  BC  =  62,500  Ib., 
DE  =  31,250  Ib.,  CX>  =  121,875  Ib.,  .EJF  =  40,640  Ib. 


INDEX 


Allowable  stresses,  7 

Allowance  for  shrinkage  and  forced  fits, 

130 
Annealing,  6 

Bach's  formula  for  flat  plates,  145 
Barlow's  formula  for  thick  cylinders, 

125 
Beams,  built-in,  80 

continuous,  70 

deflection  of,  60 

design  of,  52 

fundamental  formula  for,  51 

restrained,  80 

strength  of,  52 

Bending  and  torsion,  combined,  110 
Bending  moment,  36 
Bending  stress,  49 

Bending-rnoment  and  shear  diagrams,  37 
Bernoulli's  assumption,  50 
Birnie's  formula  for  thick  cylinders,  126 
Boiler  shells,  148 
Built-in  beams,  80 

Cantilever  beams,  60 
Cantilevers,  deflection  of,  63 
Cement,  natural,  157 

Portland,  157 

standard  mixtures  of,  158 
Center  of  curvature,  60 

of  gravity,  17 

of  mass,  19 
Centroid,  19 

of  circular  arc,  21 

of  circular  sector,  22 

of  circular  segment,  22 

of  composite  figures,  24 

of  parabolic  segment,  23 

of  triangle,  20 
Clapeyron,  70 
Clavarino's  formula  for  thick  cylinders, 

126 

Coefficient  of  linear  expansion,  10 
Column  formulas,  comparison  of,  99 
Columns,  91 

eccentrically  loaded,  101 

reenforced-concrete,  164 
Compression,  2 
Concrete,  reenforced,  156 
Conditions  of  equilibrium,  35 


Continuous  beams,  70 

Considered  formula  for  column  hoops, 

166 

Cooper's  column  formulas,  101 
Cottered  joints,  14 
Couple,  definition  of,  177 
Curvature  of  beams,  60 
Cylinders,  built  up,  126 

rational  formulas  for,  127 

thick,  stress  in,  120-126 

thin,  stress  in,  119 

Deflection  of  beams,  60 

angular,  61,  62 
Deformation,  unit,  4 

Eccentrically  loaded  columns,  101 
Efficiency  of  riveted  joints,  146 
Elastic  curve,  50,  60 
Elastic  limit,  5 
Elastic  resilience,  8 

Equilibrium,  conditions  of,  35,  178,  181 
Equilibrium  polygon,  179,  181-185 
Euler's  column  formula,  92 
Expansion,  coefficient  of  linear,  10 

Factor  of  safety,  6 

Fatigue  of  metals,  5 

Flat  plates,  empirical  formulas  for,  145 

stress  in,  136-143 

theory  of,  136 

Flat  slabs,  reenforced-concrete,  167,  173 
Forced  fits,  129 
Forces,  composition  and  resolution  of,  176 

Graphical  methods,  187,  189,  192 
Grashof  s  formula  for  flat  plates,  145 
Guest's  formula  for  shafts,  109 

Hooke's  law,  4 
Hoop  stress,  118 
Horse-power  formula,  114 

Impact,  9 

Johnson's  parabolic  column  formula,  97 
Johnson's  straight-line  column  formula, 

99 

Joint  reactions,  calculation  of,  188 
Joints,  riveted,  146 


209 


210 


RESISTANCE  OF  MATERIALS 


Lamp's  formulas  for  thick   cylinders, 

120 
Lime,  157 

Maxwell  diagrams,  190 
Modulus  of  elasticity,  4 

of  resilience,  8 

of  rigidity,  5 

of  shear,  5 

Moment,  definition  of,  15,  176 
Moment  of  inertia,  25 

for  circle,  28 

for  composite  figures,  29 

for  rectangle,  26 

for  triangle,  28 

theorems  on,  27 
Moment  of  resistance,  52 
Moment  and  shear  diagrams,  37 

directions  for  sketching,  44 

properties  of,  43 

relations  between,  41 
Moment  solid,  26 
Moments,  in  continuous  beams,  77 

fundamental  theorem  of,  15 
Mushroom      system      of     reenforced- 
concrete  construction,  167 

Neutral  axis,  50 

Nichols's  formula  for  flat  plates,  145 

Poisson's  ratio,  9 

Portland  cement,  156 

Power  transmitted  by  shafts,  108 

Quicklime,  156 

Radially  reenforced  flat  slabs,  167 
Rankine's  column  formula,  95 
Rankine's  formula  for  shafts,  109 
Reenforced  concrete,  156 
Reenforced-concrete  beams,  design  of, 

159 

Reenforced-concrete  columns,  164 
Reenforcement,  calculation  of,  in  reen- 

forced-concrete  beams,  162 
Resilience,  8 

of  shafts,  110 
Resistance,  moment  of,  52 
Restrained  beams,  80 
Resultant,  16 
Rivet  pitch,  150 
Riveted  joints,  146 

efficiency  of,  146 


Settlement  of  supports,  effect  of,  76 
Shafts,  angle  of  twist  in,  107,  112 

elliptical,  111 

power  transmitted  by,  108 

rectangular,  111 

resilience  of,  110 

square,  111 

stress  in,  106 

triangular,  112 
Shear,  2 

angle  of,  106 

in  continuous  beams,  78 

vertical,  36 
Shear  modulus,  5 
Shrinkage  and  forced  fits,  129 
Simple  beams,  deflection  of,  65 
Slag  cement,  158 

Specifications  for  structural  steel,  150 
Spheres,  thin,  stress  in,  118 
Spider,  dimensions  of,  174 
Spider  hoops,  168 
Static  moment,  15 
Stirrups  in  reenforced-concrete  beams, 

162 

Straight-line  law,  50 
Strain,  definition  of,  1 

varieties  of,  2 
Strain  diagram,  4 
Stress,  definition  of,  1 

in  beams,  50 

unit,  4 

Stresses,  in  structures,  185-194 
Structural-steel  riveting,  149 
Structures,  stresses  in,  176,  185-194 
Struts,  191 
Symmetry,  axis  of,  24 

Temperature  stress,  9 

Tension,  2 

Theorem  of  three  moments,  70 

Three  moments,  theorem  of,  70 

Thurston's  formula  for  flat  plates,  145 

Torsion,  3,  106 

Twist,  angle  of,  106 

Ultimate  strength,  5 

Unit  deformation,  4 

Unit  stress,  4 

Unit  stresses  in  structural  steel,  150 

Web     reenforcement     in     reenforced- 
concrete  beams,  162 
Working  stress,  6 


Section  modulus,  52 
Sections,  method  of,  190 


Yield  point,  5 

Young's  modulus  of  elasticity,  4 


ANNOUNCEMENTS 


TEXT-BOOK  ON  THE  STRENGTH  OF 

MATERIALS  (Revised  Edition) 

By  S.  E.  SLOCUM,  Assistant  Professor  of  Mathematics  in  the  University  of  Illinois,  and 
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